SGCh16 - Chapter Sixteen ACID-BASE EQUILIBRIA AND...

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Unformatted text preview: Chapter Sixteen ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA • • • • • • Buffer Solutions Titration Curves and Indicators Solubility and Solubility Product Predicting Precipitation Reactions and Separation of Ions by Precipitation Factors Affecting Solubility: Common Ion Effect and pH Complex Ions and Solubility BUFFER SOLUTIONS STUDY OBJECTIVES 1. 2. 3. 4. Describe the effect of common ions on the percent ionization of weak acids and bases. Calculate the pH of a buffer solution, given the concentrations of weak acid or base and their salts. Determine the pH of a buffer solution after the addition of a small amounts of strong acid base. Describe how to prepare a buffer solution with a specific pH. The Common Ion Effect. A 1.0 M HF solution has equal concentrations of hydrogen ion and fluoride ion; [H +] = [F– ] = 2.6 × 10 –2 M. Upon dissolving enough sodium fluoride (NaF) to bring the F– ion concentration up to 1.0 M, the hydrogen ion concentration will fall to 7.1 × 10 –4 M. The effect of addition of NaF to an HF solution is an example of the common ion effect. In this equilibrium, fluoride ion is the common ion. The equilibrium in the original hydrofluoric acid solution was HF(aq) H+(aq) + F– (aq) When sodium fluoride, a strong electrolyte, was added the concentration of F– ion increased. NaF(aq) → Na+(aq) + F– (aq) According to Le Chatelier's principle, the addition of F– ions will shift the weak acid equilibrium to the left; this consumes some of the F– ions and some H +(aq), and lowers the percent ionization. In effect, the percent ionization of a weak acid (HA) is repressed by the addition to the solution of its conjugate base, F– ion. The shift in equilibrium caused by the addition of an ion common to one of the products of the original equilibrium reaction is called the c ommon ion effect . Buffer Solutions. Any solution like the preceding one, which contains both a weak acid (HF) and its conjugate base (F– ) has the ability to neutralize added strong acids and bases with very little change in its pH. Such a solution is called a b uffer s olution because it resists significant changes in the pH. A buffer must contain an acid to react with any OH– ions that may be added, and a base to react with any added H+ ions. A 3 29 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 30 / Acid-Base Equilibria and Solubility Equilibria solution containing (1) a weak acid or weak base and (2) its salt is a buffer solution. H+(aq) + F– (aq) HF(aq) Added substances that form the buffer solution When a small amount of a strong acid is added to the buffer, it is neutralized by the weak base. Also, when a small amount of a strong base is added to a buffer, it is neutralized by the weak acid. In the hydrofluoric acidfluoride ion buffer, for example, the following neutralization reactions take place: • On addition of a strong base, the neutralization reaction is OH – + H F → H2 O + F – • On addition of a strong acid, the neutralization reaction is H + + F – → HF In order to be most effective, the amounts of weak acid and weak base used to prepare the buffer must be considerably greater than the amounts of strong acid or strong base that may be added later. B uffer capacity refers to the amount of acid or base that can be neutralized by a buffer solution before its pH is appreciably affected. The pH of a Buffer. To calculate the pH of a buffer requires that the concentrations of the weak acid (such as HF) and a soluble salt of the weak acid (NaF) be substituted into the ionization constant expression for the weak acid. H+(aq) + F– (aq) HF(aq) [H+][F– ] [HF] +] = [HF] K [H a [F– ] Ka = The pH is found from pH = – log [H+]. Alternatively, the Henderson-Hasselbalch equation, which is introduced in the textbook in Section 16.3, can be used to calculate buffer pH. pH = pKa + log [conjugate base] [acid] where the pKa = – log Ka. This equation holds only if the acid is less than 5 percent ionized. This requirement is almost always met because the ionization of the weak acid is lowered by the addition of the conjugate base in accord with the common ion effect. Use of the Henderson-Hasselbalch equation is shown in Example 16.1. Preparing a Buffer. Each weak acid–conjugate base buffer has a characteristic pH range over which it is most effective. In terms of the Henderson-Hasselbalch equation, the pH of a buffer depends on the pKa of the weak acid, and the ratio of conjugate base concentration to weak acid concentration. A buffer has equal ability to neutralize either added acid or added base when the concentrations of its weak acid and conjugate base components are equal. When [conj base] / [weak acid] = 1, then log [conjugate base] =0 [acid] and the Henderson-Hasselbalch equation becomes pH = pKa Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Acid-Base Equilibria and Solubility Equilibria / 3 31 In order to prepare a buffer of a certain pH, the weak acid must have a pKa as close as possible to the desired pH. Next we substitute the desired pH and the pKa of the chosen acid into the Henderson-Hasselbalch equation in order to determine the ratio of [conj base] to [weak acid]. The ratio can then be converted into molar quantities of the weak acid and its salt. _______________________________________________________________________________ EXAMPLE 16.1 pH of a Buffer Solution Calculate the pH of a buffer solution that is 0.25 M HF and 0.50 M NaF. •Method of Solution The pH of a HF/F– buffer is given in terms of the Henderson-Hasselbalch equation. HF(aq) H+(aq) + F– (aq) pH = pKa + log [F– ] [HF] where pKa = – log Ka pKa = – log (7.1 × 10 –4 ) = 3.15 The equilibrium concentrations of F– ion and HF are determined as follows. The initial concentration of HF is 0.25 M. But at equilibrium [HF] = 0.25 – x. We can assume that x is less than 5 percent of [HF]0 because HF is a weak acid, and because the addition of F– ion, a common ion, represses the dissociation of HF. We assume that [HF] = [HF]0 = 0.25 M. The F – ion is contributed by two sources, sodium fluoride which is a strong electrolyte, and HF, a weak acid. NaF(aq) → Na+(aq) + F– (aq) Therefore, [F– ] = 0.50 M + x. Since 0.50 M >> x, then [F– ] = 0.50 M. Substituting into the Henderson-Hasselbalch equation gives: pH = 3.15 + log 0.50 = 3.15 + log 2.0 = 3.15 + 0.30 0.25 = 3.45 _______________________________________________________________________________ EXAMPLE 16.2 Percent Ionization in a Buffer Solution Calculate the percent ionization of HF in the buffer solution of the preceding example. •Method of Solution percent ionization = where [H+] × 100% [HF]0 [H+] = 10–pH = 10 –3.45 [H+] = 3.55 × 10 –4 M percent ionization = Back Forward Main Menu 3.55 × 1 0 –4 M 0.25 TOC × 100% = 0.14% Study Guide TOC Textbook Website MHHE Website 3 32 / Acid-Base Equilibria and Solubility Equilibria •Comment By comparison, the percent ionization of HF in a 0.25 M HF solution is 5.3%. The common ion effect lowers the percent ionization to the value found in the buffer solution. Neglecting x was a valid assumption. _______________________________________________________________________________ EXAMPLE 16.3 Adding Strong Acid to a Buffer Suppose 3.0 mL of 2.0 M HCl is added to exactly 100 mL of the buffer described in Example 16.1. What is the new pH of the buffer after the HCl is neutralized? •Method of Solution The HCl is neutralized in this buffer by the following reaction which goes to completion. H+ + F – → HF This consumes some F– ions and forms more HF. The number of moles of H+ added as HCl is M·V = 2.0 mol/L × 0.0030 L = 0.0060 mol H+ The number of moles of HF originally present in 100 mL of buffer was M·V = 0.25 mol/L × 0.100 L = 0.025 mol HF The number of moles of F– originally present in 100 mL of buffer was M·V = 0.50 mol/L × 0.100 L = 0.050 mol F– •Calculation After the added H+ is neutralized by H+ + F – → HF, 0.0060 H+ reacts with 0.0060 mol of F– ion to form 0.0060 mol HF. The number of moles of HF is 0.025 + 0.006 = 0.031 mol. The number of moles of F– is 0.050 – 0.006 = 0.044 mol. The new pH can be found as usual from [F– ] [HF] (0.044 mol/0.103 L) = 3.15 + log = 3.15 + log 1.42 = 3.15 + 0.15 (0.031 mol/0.103 L) pH = pKa + log = 3.30 •Comment The pH has dropped only 0.15 unit from 3.45 (Example 16.1) due to the addition of 3.0 mL of 2.0 M HCl; thus the use of the term buffer solution. _______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Acid-Base Equilibria and Solubility Equilibria / 3 33 _______________________________________________________________________________ EXAMPLE 16.4 Conjugate Base to Weak Acid Ratio in a Buffer What ratio of [F– ] to [HF] would you use to make a buffer of pH = 2.85? •Method of Solution Start with the Henderson-Hasselbalch equation. pH = pKa + log [conjugate base] [acid] • Calculation Substitute in the desired pH and the pKa value of HF. 2.85 = 3.15 + log [F– ] [HF] [F– ] = – 0.30 [HF] [F– ] = 10 –0.30 = 0.50 [HF] log The ratio of [F – ] to [HF] should be 0.50 to 1. Any amounts of F– and HF that give a ratio of 0.5 will produce the desired pH. _______________________________________________________________________________ EXERCISES 1. 2. a. What is the pH of a buffer solution prepared by dissolving 0.15 mol of benzoic acid (C6 H5 COOH) and 0.45 mol of sodium benzoate (C6 H5 COONa) in enough water to make 400 mL of solution? b. Does the volume of solution affect your answer? 3. What mole ratio of benzoate ion to benzoic acid would you need to prepare a buffer solution with a pH of 5.00? 4. a. What is the change in pH when 40.0 mg of NaOH is added to 100 mL of a buffer solution consisting of 0.165 M NH3 and 0.120 M NH 4 Cl? b. Write the equation for the buffer equilibrium. c. Write the equation for the neutralization of NaOH by the buffer. 5. Back Compare the percent ionization of acetic acid in 0.10 M CH 3 COOH to that in a solution of 0.020 M CH3 COONa and 0.10 M CH 3 COOH. What is the optimum pH of a H3 PO4 / H 2 PO4 buffer? Forward – Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 34 / Acid-Base Equilibria and Solubility Equilibria TITRATION CURVES AND INDICATORS STUDY OBJECTIVES 1. 2. Describe the shape of titration curves for titrations involving a strong acid and a strong base, a weak acid and a strong base, and a strong acid and a weak base. Choose the best indicator for a particular acid-base titration. Titration Curves. Acid-base titrations were discussed in Chapter 4. With the introduction to pH in the previous chapter, it is informative to follow the pH as a function of the progress of the titration. A graph of pH versus volume of titrant added is called a titration curve. Initially, the pH is that of the unknown solution. As titrant is added, the pH becomes that of a partially neutralized solution of unknown plus titrant. The pH at the equivalence point refers to the [H+] when just enough titrant has been added to completely neutralize the unknown. If more titrant is added after the equivalence point has been reached, the pH assumes a value consistent with the pH of excess titrant. Titration curves are shown in the textbook. Figure 16.3 of the text shows the titration of a strong acid by the addition of a strong base. The main features of this curve can be stated briefly. The pH starts out quite low. As base is slowly added, it is neutralized, and the pH is determined by the unreacted acid. Near the equivalence point, the pH begins to rise more rapidly. At the equivalence point the pH changes rapidly (about 5.0 units upon the addition of only two drops of base). Beyond the equivalence point the pH is determined by the amount of excess base that is added. The pH at the equivalence point of an acid-base titration is the pH of the salt solution that is formed by neutralization. NaCl, NaNO3 , NaBr, KCl, and KI are examples of salts that can be formed in titrations of strong acids and strong bases. These salts yield ions that do not cause hydrolysis. Therefore, the equivalence point in a titration of a strong acid with a strong base occurs at pH 7.0. Figure 16.5 of the text shows titration curves for a strong acid with a weak base and for a weak acid with a strong base. In the titration of a strong acid with a weak base, the first part of the curve is the same as in the strong acid versus strong base titration. However, the pH at the equivalence point is below 7.0 because the cation of the salt is the conjugate acid of the weak base used in the titration. Hydrolysis of this salt yields an acidic solution. In the titration of hydrochloric acid with ammonia, the salt produced is NH4 Cl. The Cl– ion does not hydrolyze, but NH+ does: 4 NH+ + H 2 O 4 NH3 + H3 O+ In the titration of a weak acid with a strong base, the initial pH is greater than in the titration of a strong acid (Figure 16.4 text). At the equivalence point, the pH is above 7.0 because of salt hydrolysis. The anion of the salt is the conjugate base of the weak acid used in the titration. In the titration of acetic acid with sodium hydroxide the salt produced is CH3 COONa. The Na+ ion does not hydrolyze, but CH 3 COO– does: CH3 COO– + H2 O CH3 COOH + OH – This makes the solution basic at the equivalence point. Indicators. Indicators are used in the laboratory to reveal the equivalence point of a titration. The abrupt change in color of the indicator signals the endpoint of a titration which usually coincides very nearly with the equivalence point. Indicators are usually weak organic acids that have distinctly different colors in the nonionized (molecular) and ionized forms. [H+][In– ] HIn(aq) H+(aq) + In– (aq) Ka = [HIn] color 1 color 2 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Acid-Base Equilibria and Solubility Equilibria / 3 35 To determine the pH range in which an indicator will change color we can write the Ka expression in logarithmic form. pH = pKa + log [In– ] [HIn] The color of an indicator depends on which form predominates. Typically, when [In– ]/[HIn] ≥ 10 the solution will be color 2, and when [In– ]/[HIn] ≤ 0.1, the solution will be color 1. Thus the color change will occur between pH = pKa + log (10/1) = pKa + 1 .0 and pH = pKa + log (1/10) = pKa – 1.0 At the midpoint of the pH range over which the color changes, [HIn] = [In– ], and pH = pKa. Like any weak acid each HIn has a characteristic pKa, and so each indicator changes color at a characteristic pH. Table 16.1 of the text lists a number of indicators used in acid-base titration and the pH ranges over which they change color. The choice of indicator for a particular titration depends on the expected pH at the equivalence point. For instance, in the titration of acetic acid by sodium hydroxide, which is discussed in Example 16.7 of the text, the pH at the equivalence point is 8.72. According to Table 16.1 both cresol red and phenolphthalein change color over ranges that include pH 8.72. Therefore either of these indicators would show the equivalence point of this titration. _______________________________________________________________________________ EXAMPLE 16.5 Net Ionic Equations Write the net ionic equations for the neutralization reactions that occur during the following titrations. Predict whether the pH at the equivalence point will be above, below, or equal to 7.0. a. Titration of HI with NH 3 . b. Titration of HI with NaOH. c. Titration of NaOH with HF. •Method of Solution a. This is a titration of a strong acid with a weak base. HI is 100 percent dissociated, and NH3 is a weak base. The net ionic equation for neutralization is + H+ + NH 3 → NH4 + + The result of the neutralization reaction is formation of the conjugate acid NH4 of the weak base NH3 . NH 4 i s an acid. Answer: The pH at the equivalence point will be below 7.0. b. This is a titration of a strong acid and a strong base. The net ionic equation is H+ + OH – → H2 O The resulting solution is neutral at the equivalence point. Answer: pH = 7.0 c. This titration involves a weak acid and a strong base. HF + OH– → H2 O + F – The reaction produces the conjugate base (F– ion) of the weak acid. Answer: The solution will be basic at the equivalence point, pH > 7.0. _______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 36 / Acid-Base Equilibria and Solubility Equilibria _______________________________________________________________________________ EXAMPLE 16.6 Choosing an Indicator Choose an indicator for the titration of 50 mL of a 0.10 M HI solution with 0.10 M NH3 . HI(aq) + NH3 (aq) → NH4 I(aq) •Method of Solution First we need to know the pH at the equivalence point. This is the titration of a strong acid with a weak base. The net ionic equation is + H+(aq) + NH3 (aq) → N H 4 (aq) –3 m ol + 5.0 × 10 –3 mol yields 5.0 × 10 –3 mol 5.0 × 10 The product NH+ ion is a weak acid and so we expect a slightly acidic solution at the equivalence point. 4 •Calculation + + The concentration of NH4 formed at the equivalence point is 5.0 × 10 –3 mol/0.100 L = 5.0 × 10 –2 M NH 4 . + The [H+] at the equivalence point is just the pH of a 0.050 M NH4 solution. Consider the ionization of the + weak acid, NH4 : + NH4 (aq) NH3 (aq) + H+(aq) Ka = 5.6 × 10 –10 _________________________________________________ + Concentration NH4 H+ + NH3 _________________________________________________ Initial (M) 0.050 0 0 Change (M) –x +x +x ___________________________ Equilibrium (M) (0.050 – x) x x _________________________________________________ Ka = [NH3 ][H+] + [NH4 ] = x2 x2 = = 5.6 × 10 –10 0.050 (0.050 – x ) Solving for x: x = [NH3 ] = [H+] = 5.3 × 10 –6 M and the pH is 5.28 at the equivalence point. According to Table 16.1 of the text, chlorophenol blue and methyl red are indicators that will change color in the vicinity of pH 5.28. _______________________________________________________________________________ EXERCISES 6. 7. Back Approximately what range of pH should be expected at the equivalence points in the titration of weak acids with strong bases? The pH at the equivalence point in a titration was found to be approximately 5. Describe the category of titration in terms of the acid and base strengths. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Acid-Base Equilibria and Solubility Equilibria / 3 37 8. 25.0 mL of 0.222 M HBr was titrated with 0.111 M NaOH. a. Write the overall balanced equation for reaction. b. After adding 30.0 mL of base solution, what is the pH? 9. Calculate the pH in the following titration after the addition of 12.0 mL of 0.100 M KOH to 20.0 mL of 0.200 M CH3 COOH. 10. Bromothymol blue is a typical acid-base indicator. It has a Ka = 1.6 × 10 –7 . Its undissociated form HIn is yellow, and its conjugate base In– is blue. What color would a solution have at pH = 5.6? 11. Bromothymol blue is a typical acid-base indicator. It has a Ka = 1.6 × 10 –7 . At what pH does this indicator change color? SOLUBILITY AND SOLUBILITY PRODUCT STUDY OBJECTIVES 1. 2. 3. Write the solubility product expression for an insoluble salt. Calculate the solubility product constant of a salt given its solubility. Calculate the solubility of a salt given its Ksp . The Solubility Product Constant. The solubility rules were discussed in Chapter 4. In this chapter, solubility is treated quantitatively in terms of equilibrium. In a saturated solution of silver bromide, for example, the solubility equilibrium is AgBr(s) Ag+(aq) + Br– (aq) The equilibrium constant for the reaction in which a solid salt dissolves is called the solubility product constant, Ksp . Ksp = [Ag+][Br– ] The solubility product expression is always written in the form of the equilibrium constant expression for the solubility reaction. Two additional examples are Ca(OH)2 and Ag2 CrO4 : Ca(OH)2 (s) Ca 2+ (aq) + 2OH– (aq) Ksp = [Ca2+ ][OH– ]2 and Ag2 CrO4 (s) 2– 2Ag+(aq) + CrO4 (aq) 2– Ksp = [Ag+]2 [CrO4 ] The solubility product constant (also called a Ksp value) can be calculated by substituting the concentrations of the ions in a saturated solution into the solubility product expression. Take AgBr again as an Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 38 / Acid-Base Equilibria and Solubility Equilibria example. Since AgBr is a strong electrolyte, all of the solid AgBr that dissolves is dissociated into Ag+ and Br– ions, AgBr(aq) → Ag+(aq) + Br– (aq) The solubility of AgBr at 25°C is 8.8 × 10 –7 moles per liter of solution. Therefore, the ion concentrations in a saturated AgBr solution are [Ag+] = [Br– ] = 8.8 × 10 –7 M The solubility product constant is found by substituting these concentrations into the solubility product: Ksp = [Ag+][Br– ] = (8.8 × 10 –7 ) (8.8 × 10 –7 ) Ksp = 7.7 × 10 –13 This calculation points out the difference between the solubility and the Ksp value. These two quantities are not the same. Here we see that the solubility is used to determine the ion concentrations, and that substitution of these into the solubility product expression gives the solubility product constant (Ksp value). Table 16.2 in the text and Table 16.1 in this chapter list solubility product constants for a number of slightly soluble salts, including silver bromide. The Ksp value can be used to calculate the solubility of a compound in moles per liter (the molar solubility), or in grams per liter. For example, the Ksp value for CaSO4 is 2.4 × 10 –5 . What is the solubility of CaSO4 in mol/L? Start with the solubility equilibrium: CaSO4 (s) 2– Ca 2+ (aq) + SO4 (aq) 2– Ksp = [Ca2+ ][SO4 ] 2– Let s equal the solubility in mol/L. At equilibrium, both [Ca2+ ] and [SO4 ] are equal to s because when s 2– moles of CaSO 4 dissolve in 1 L, s moles of Ca2+ ions and s moles of SO4 ions are produced. Substitute into the Ksp expression: 2– Ksp = [Ca2+ ][SO4 ] = (s)(s) = s2 The solubilty product is given above: Ksp = 2.4 × 10 –5 s2 = 2.4 × 10 –5 s= 2 .4 × 1 0 –5 = 4.9 × 10 –3 M It is important to notice that the molar solubility and Ksp are not the same. Molar solubility and K sp are related but are not the same quantity. Solubility and Ksp. Table 16.1 below lists a number of solubility product constants for various salts that will be needed in this chapter. Notice that within a series of salts of the same type of formula (MX, MX2 , etc.), the solubility increases as the solubility product constant increases. Comparing the Ksp 's of the three MX salts, we can see that CaSO4 has the highest Ksp value. Therefore, its solubility equilibrium lies farthest to the right of the three salts, and it is the most soluble. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Acid-Base Equilibria and Solubility Equilibria / 3 39 Table 16.1 Solubility Products of Some Slightly Soluble Salts at 25°C ___________________________________________________________ Name Formula Ksp ___________________________________________________________ MX type ___________________________________________________________ Calcium sulfate CaSO4 2.4 × 10 –5 Silver chloride AgCl 1.6 × 10 –10 Silver bromide AgBr 7.7 × 10 –13 ____________________________________________________________ MX2 and M2 X type ____________________________________________________________ Lead chloride PbCl 2 2.4 × 10 –4 Calcium hydroxide Ca(OH)2 8.0 × 10 –6 Lead iodide PbI2 1.4 × 10 –8 Magnesium fluoride MgF 2 6.4 × 10 –9 Magnesium hydroxide Mg(OH)2 1.2 × 10 –11 Silver chromate Ag2 CrO4 1.1 × 10 –12 _____________________________________________________________ _______________________________________________________________________________ EXAMPLE 16.7 Calculating a K s p V alue The solubility of magnesium fluoride in water is 7.3 × 10 –3 g per 100 mL of solution. What is the solubility product constant for MgF2 ? •Method of Solution Write the solubility equilibrium and the K sp expression: MgF 2 (s) Mg 2+ (aq) + 2F– (aq) Ksp = [Mg2+ ] [F– ]2 Calculate s, the molar solubility of MgF2 , from the given solubility. The molar mass of MgF2 is 62.31 g/mol. s= 7.3 × 1 0 –3 g 100 mL × 1 mL 1 m ol × = 1.17 × 10 –3 mol/L –3 L 62.31 g 10 The Ksp value can be calculated by substituting the ion concentrations in a saturated solution into the solubility product expression. When s mol of MgF2 dissolves in 1 L of solution, the Mg2+ concentration is equal to s, and the F– ion concentration is 2s. [Mg2+ ] = s = 1.17 × 10 –3 M [F– ] = 2s = 2(1.17 × 10 –3 M) = 2.34 × 10 –3 M • Calculation Now substitute the ion concentrations into the Ksp expression. Ksp = (1.17 × 10 –3 ) (2.34 × 10 –3 )2 Ksp = 6.4 × 10 –9 _______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 40 / Acid-Base Equilibria and Solubility Equilibria _______________________________________________________________________________ EXAMPLE 16.8 Calculating a Molar Solubility The Ksp value for Ag2 CrO4 is 1.1 × 10 –12 . Calculate the molar solubility of silver chromate. •Method of Solution Write the solubility equilibrium Ag2 CrO4 (s) 2– 2Ag+(aq) + CrO4 (aq) and the Ksp expression. 2– Ksp = [Ag+]2 [CrO4 ] Let s = molar solubility of Ag2 CrO4 . Whenever s moles of Ag 2 CrO4 dissolve, 2s moles of Ag+ and s moles of 2– CrO4 are produced. [Ag+] = 2s 2– [CrO4 ] = s Summarize the changes in concentrations as follows. ___________________________________________________ 2– Concentration Ag 2 CrO4 (s) 2Ag+(aq) + CrO4 (aq) ___________________________________________________ Initial (M) 0 0 Change (M) 2s s _________________________________ Equilibrium (M) 2s s ___________________________________________________ •Calculation Substitute these values into the Ksp expression. Ksp = [2s]2 [s] = 1.1 × 10 –12 and solve for s. 4s2 (s) = 1.1 × 10 –12 4 s 3 = 1.1 × 10 –12 3 3 1 .1 × 1 0 –12 s= = 2.75 × 1 0 –13 4 s = 6.5 × 10 –5 M _______________________________________________________________________________ EXERCISES 12. Write balanced equations and solubility product expressions for the solubility equilibria of the following compounds. a. BaCO3 b . CaF 2 c. Al(OH)3 d. Ag3 PO4 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Acid-Base Equilibria and Solubility Equilibria / 3 41 13. The solubility of PbBr2 is 0.392 g per 100 mL at 20°C. What is the Ksp value for PbBr2 ? 14. a. What is the solubility of CaF2 in moles per liter given that Ksp = 4.0 × 10 –11 . b. Determine the F– ion concentration in a saturated solution of CaF2 . 15. Solid barium fluoride was dissolved in pure water until a saturated solution was formed. If the fluoride concentration was 0.0150 M, what the Ksp of BaF2 ? PREDICTING PRECIPITATION REACTIONS AND SEPARATION OF IONS BY PRECIPITATION STUDY OBJECTIVES 1. 2. 3. Predict whether a precipitate will form when two solutions of salts are mixed. Describe how to selectively precipitate one of two ions in a solution. Determine the completeness of precipitation. Criteria for Precipitate Formation. When two solutions containing dissolved salts are mixed, formation of an insoluble compound is always a possibility. From a knowledge of solubility rules (Chapter 4) and solubility products (Table 16.1), you can predict whether a precipitate will form. For a dissolved salt MX(aq), the ion product (Q) is Q = [M+]0[X– ]0, where [ ] 0 stands for the initial concentrations. Any one of three conditions for the ion product Q may exist after two solutions are mixed. 1. 2. 3. Q = Ksp For a saturated solution, the value of Q is equal to Ksp . No precipitate will form in this case as no net reaction occurs in a system at equilibrium. Q < K sp In an unsaturated solution the value of Q is less than Ksp . No precipitate will form in this case. Q > K sp In a supersaturated solution more of the salt is dissolved than the solubility allows. In this case, an unstable situation exists, and some solute will precipitate from solution until a saturated solution is attained. At this point Q = Ksp and equilibrium is reestablished. Suppose 500 mL of a solution containing 2.0 × 10 –5 M Ag + is mixed with 500 mL of solution containing 2.0 × 10 –4 M Br– . Will a precipitate form? First determine the new concentrations of Ag+ and Br– in the mixture. The total volume is 1 L, and so accounting for dilution the ion concentrations in the new solution are just half of what they were in the separate solutions. The initial concentrations before any precipitate forms are: [Ag+]0 = 1 × 10 –5 M [Br– ]0 = 1.0 × 10 –4 M The ion product for AgBr is Q = [Ag+]0 [Br– ]0 To predict whether a precipitate of AgBr will form, calculate the ion product Q, and compare it to the Ksp value. Q = [Ag+]0 [Br– ]0 = (1 × 10 –5 )(1 × 10 –4 ) = 1 × 10 –9 Ksp = 7.7 × 10 –13 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 42 / Acid-Base Equilibria and Solubility Equilibria Therefore, Q > Ksp , and so the mixture corresponds to a supersaturated solution of AgBr, which means that some AgBr(s) will precipitate: Ag+(aq) + Br– (aq) → AgBr(s) until a new equilibrium is reached at which point Q = K sp . Criteria for Separation of Ions by Precipitation. Ions in a solution can be separated from each other on the basis of the different solubilities of their salts. For example, when slowly adding a solution containing OH– ion to a solution that contains both Ca2+ ions and Mg 2+ ions, we find that Mg(OH) 2 precipitates before Ca(OH)2 . Mg(OH)2 with a Ksp of 1.2 × 10 –11 precipitates first because it is less soluble than Ca(OH)2` whose Ksp is 8.0 × 10 –6 . The concentrations of the ions in solution also help determine which ions will precipitate first. Even though Mg(OH)2 is less soluble than Ca(OH)2 , Ca(OH)2 would precipitate first when OH– ion is added if the concentration of Ca2+ ion was much greater than the concentration of Mg2+ ion. It is best to determine the concentration of OH– ion needed to cause a precipitate of both compounds. Since Ca2+ and Mg2+ ion concentrations will be given, then the concentration of OH – that must be exceeded to initiate the precipitation of Ca(OH)2 and Mg(OH)2 are, respectively: [OH– ] = Ksp [Ca2+ ] and [OH– ] = Ksp [Mg2+ ] 2– Examples 16.10 and 16.11 illustrate the separation of Cl– and CrO4 (chromate) ions. _______________________________________________________________________________ EXAMPLE 16.9 Predicting Formation of a Precipitate a. b. Predict whether or not a precipitate of PbI2 will form when 200 mL of 0.015 M Pb(NO3 )2 and 300 mL of 0.050 M NaI are mixed together. Given K sp (PbI2 ) = 1.4 × 10 –8 . If the answer is yes, what concentrations of Pb 2+ (aq) and I– (aq) will exist when equilibrium is reestablished? •Method of Solution for (a) Recall that Pb(NO3 )2 and NaI are both strong electrolytes. When the solutions are mixed, will the following reaction occur? – – Pb 2+ (aq) + 2NO3 (aq) + 2Na+(aq) + 2I– (aq) → PbI2 (s) + 2Na+(aq) + 2NO3 (aq) A precipitate will form only if the ion product exceeds the Ksp value; [Pb 2+ ]0 [I– ]2 > K sp . 0 When the two solutions are mixed, 500 mL of new solution is formed. Immediately after mixing, the initial ion concentrations would be: 2 00 mL = 6.0 × 10 –3 M 500 mL 3 00 mL [I– ]0 = 0.050 M × = 3.0 × 10 –2 M 500 mL [Pb2+ ]0 = 0.015 M × Then Q = [Pb2+ ]0 [I– ]2 = (6.0 × 10 –3 )(3.0 × 10 –2 )2 = 5.4 × 10 –6 0 Since Q > Ksp , then a precipitate of PbI 2 will form. •Method of Solution for (b) The concentrations of Pb2+ and I– remaining in solution depend on the number of moles initially and the number of moles that precipitate. The initial number of moles of Pb2+ ion was 0.015 M × 0.200 L = 0.0030 mol, and that of I – ion, 0.050 M × 0.300 L = 0.015 mol. In order to determine how many moles of Pb2+ and I– precipitated from solution, consider the net reaction Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Acid-Base Equilibria and Solubility Equilibria / 3 43 Pb 2+ (aq) + 2I– (aq) → PbI2 (s) Look to see if there is a limiting reagent. Then determine the amount of the excess reagent that remains after complete reaction of the limiting reagent. Pb2+ ion being in the smaller amount is the limiting reagent, and will react essentially completely. The 0.0030 mol of Pb2+ will react with 0.0060 mol of I– , leaving 0.009 mol of excess I– (0.015 – 0.0060) dissolved in solution. •Calculation The I– concentration is [I– ] = 0.009 mol/0.50 L = 0.018 M. The Pb 2+ concentration is controlled by the PbI2 solubility equilibrium. In other words, some PbI2 dissolves by the reverse reaction. Pb 2+ + 2I– PbI2 (s) Ksp = [Pb 2+ ][I– ]2 = 1.4 × 10 –8 The Pb 2+ ion concentration in equilibrium with 0.018 M I– ion is: Ksp 1 .4 × 1 0 –8 [Pb2+ ] = – 2 = = 4.3 × 10 –5 M [I ] (0.018)2 •Comment The percentage of Pb2+ ion remaining unprecipitated is 4.3 × 10–5 M ( 0.5 L) × 100% = 0.72% 0.0030 mol This confirms our assumption that essentially all the Pb2+ ion precipitated. _______________________________________________________________________________ EXAMPLE 16.10 Selective Precipitation of an Ion 2– A solution contains 0.10 M Cl– and 0.010 M CrO 4 . If AgNO 3 solution is added dropwise which will precipitate first, AgCl or Ag 2 CrO4 ? •Method of Solution The solubility product is a number that the product of the ion concentrations can never exceed at equilibrium. First write the two equilibria of interest and their solubility constants. Ag+(aq) + Cl– (aq) AgCl(s) Ag2 CrO4 (s) 2– 2Ag+ + CrO4 Ksp = [Ag+][Cl– ] = 1.6 × 10 –10 2– Ksp = [Ag+]2 [CrO4 ] = 1.1 × 10 –12 •Calculation The highest Ag + ion concentration possible in a solution of 0.10 M Cl– is [Ag+] = Back Forward 1.6 × 1 0 –10 1 .6 × 1 0 –10 = = 1.6 × 10 –9 M 0.10 [Cl– ] Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 44 / Acid-Base Equilibria and Solubility Equilibria 2– The highest Ag + ion concentration possible in a solution of 0.010 M CrO4 Ksp [Ag+] = 2– = [CrO4 ] is 1 .1 × 1 0 –12 = 1.0 × 10 –5 M 0.010 Silver chloride will precipitate before silver chromate because of the lower Ag+ ion concentration needed to produce a saturated AgCl solution. _______________________________________________________________________________ EXAMPLE 16.11 Completeness of Precipitation In the above example AgCl precipitates before Ag2 CrO4 as Ag +(aq) ion is added dropwise to the solution. What percentage of the Cl– ion in solution will have precipitated when Ag2 CrO4 just begins to precipitate? •Method of Solution As we found above, Ag2 CrO4 begins to precipitate when [Ag+] = 1.0 × 10 –5 M . •Calculation The Cl – ion concentration in equilibrium with 1.0 × 10 –5 M Ag + ion is: 1.6 × 1 0 –10 1 .6 × 1 0 –10 = = 1.6 × 10 –5 M 1.0 × 1 0 –5 M [Ag+] The percent of Cl– remaining in solution unprecipitated is: [Cl– ] = %Cl– = 1 .6 × 1 0 –5 M 0.10 M × 100% %Cl– = 0.016% Therefore, the percentage of Cl– ion precipitated is 100% – 0.016% = 99.98%. _______________________________________________________________________________ EXERCISES 16. If 1.0 mL of 1.0 × 10 –3 M Ba(NO3 )2 is added to 99.0 mL of 1 × 10 –4 M Na 2 CO3 , will BaCO 3 precipitate from this solution? Ksp (BaCO3 ) = 8.1 × 10 –9 . 17. Will a precipitate form when 250 mL of 2.8 × 10 –3 M MgCl2 solution is added to 250 mL of 5.2 × 10 –3 M NaF? Identify the precipitate if any. 18. Solid Ba(NO3 )2 is slowly dissolved in a solution of 2.5 × 10 –4 M Na 2 CO3 . At what Ba 2+ concentration will a precipitate of BaCO3 just begin to form? 19. A solution contains 1.0 × 10 –4 M Cu+ and 1.0 × 10 –3 M P b 2+ . a. If NaI solution is added dropwise to this solution which compound will precipitate first, CuI or PbI2 ? b. Give the concentrations of I – necessary to begin precipitation of each salt. c. What percentage of the initial Cu + ion concentration will remain unprecipitated when PbI2 just begins to precipitate? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Acid-Base Equilibria and Solubility Equilibria / 3 45 FACTORS AFFECTING SOLUBILITY: COMMON ION EFFECT AND pH STUDY OBJECTIVES 1. 2. Calculate the solubility of a salt in a solution containing an ion common to that salt. Predict the effect of pH on the solubility of a salt. The Common Ion Effect. The effect of adding an ion common to one already in equilibrium in a solubility reaction is to decrease the solubility of the salt. In the case of AgBr solubility: AgBr(s) Ag+(aq) + Br– (aq) The addition of either Ag+ or Br– ions will shift the equilibrium to the left, in accord with Le Chatelier's principle, thus decreasing the amount of AgBr dissolved. Ag+ ions can be added by pouring in a solution of AgNO3 . Recall that AgNO 3 is very soluble, and is a strong electrolyte. – AgNO3 (s) → Ag+(aq) + NO3 (aq) The nitrate ion will not interfere with AgBr solubility because it is not a common ion. Additional Br– ions could be supplied by adding NaBr, for instance. Sodium bromide is very soluble and is a strong electrolyte. NaBr(s) → Na+(aq) + Br– (aq) The sodium ion will not affect the solubility of AgBr because it is not a common ion. Here we see that the addition of an ion that is common to one already in the solubility equilibrium shifts the equilibrium to the left, which decreases the solubility. This is equivalent to the case of a weak acid, where the presence of a common ion decreases the percent ionization. Several calculations involving the common ion effect appear at the end of this section. pH and Solubility. The pH can affect the solubility of a solute in two ways. One of these is through the common ion effect. Consider the solubility equilibrium of an insoluble hydroxide such as Mg(OH)2 or Ca(OH)2 . Ca(OH)2 (s) Ca 2+ (aq) + 2OH– (aq) Upon the addition of NaOH, for instance, the pH of the solution will increase. The equilibrium position will shift to the left because of the added OH– ion (a common ion); therefore the solubility of Ca(OH)2 will decrease proportionately. The other case in which pH can affect solubility is when a salt contains a basic anion such as F– , CH3 COO– , or CN– . Any basic anion will react with H+ ions present in the solution and thereby affect the solubility of the salt. Take, for instance, the effect of adding a strong acid on the solubility of silver acetate. To explain this we need two reaction steps. In the first one silver acetate dissolves, and in the second the acetate ion combines with H +(aq) ions present in the system: CH3 COOAg(s) CH 3 COO– (aq) + Ag+(aq) H+(aq) + CH3 COO– (aq) → CH3 COOH(aq) Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 46 / Acid-Base Equilibria and Solubility Equilibria In the presence of H+ the concentration of acetate ion is lowered by the occurrence of the second reaction. The second reaction occurs essentially completely because it forms a weak acid. According to Le Chatelier's principle, the solubility equilibrium will shift to the right, causing more silver acetate to dissolve. – The solubilities of salts containing anions that do not hydrolyze, such as Cl – , Br– , I – , and NO3 , are not affected by pH. These anions an conjugate bases of strong acids. _______________________________________________________________________________ EXAMPLE 16.12 Common Ion Effect What is the solubility of PbCl2 in 0.50 M NaCl solution? •Method of Solution Write the solubility equilibrium reaction for PbCl2 , and write the Ksp expression. The K sp value is in Table 16.1. PbCl 2 (s) Pb 2+ (aq) + 2Cl– (aq) Ksp = [Pb 2+ ][Cl– ]2 = 2.4 × 10 –4 All of the [Pb2+ ] ion comes from the dissolution reaction. Let s = molar solubility of PbCl2 . The only source of Pb 2+ is PbCl2 . Therefore: [Pb2+ ] = s Chlorine ions are contributed by PbCl2 and by NaCl (a strong electrolyte): NaCl(aq) → Na+(aq) + Cl– (aq) therefore [Cl– ] = 0.50 M + 2s A table here will help summarize the changes in concentrations. ___________________________________________________ Concentration PbCl 2 (s) Pb 2+ (aq) + 2Cl– (aq) ___________________________________________________ Initial (M) 0 0.50 Change (M) s 2s ______________________________________ Equilibrium (M) s (0.50 + 2s) ____________________________________________________ •Calculation Substituting into Ksp , we get: [Pb2+ ][Cl– ]2 = Ksp (s)(0.50 + 2s) 2 = 2.4 × 10 –4 If we assume 0.50 >> 2s, then solving the problem is greatly simplified: (s)(0.50)2 = 2.4 × 10 –4 s = 9.6 × 10 –4 M •Comment Note that the assumption 0.50 >> 2s was valid; 0.50 >> 1.9 × 10 –3 M . _______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Acid-Base Equilibria and Solubility Equilibria / 3 47 _______________________________________________________________________________ EXAMPLE 16.13 Solubility of a Metal Hydroxide What is the solubility of Pb(OH)2 in a buffer solution of pH 8.0 and in another of pH 9.0? Given Ksp = 1.2 × 10 –15 •Method of Solution Write the solubility equilibrium and the K sp expression for Pb(OH)2 . Pb(OH)2 (s) Pb 2+ (aq) + 2OH– (aq) Ksp = [Pb 2+ ][OH– ]2 = 1.2 × 10 –15 Let s equal the molar solubility of Pb(OH)2 . The concentration of Pb 2+ will be equal to s. The concentration of OH– ion will not be 2[Pb2+ ] because in a buffer solution, the H + ion and OH– ion concentrations are maintained constant by the buffer. At pH 8.0, the pOH is 6.0, and [OH– ] = 1.0 × 10 –6 M. In the buffer solution, the [OH– ] ion concentration is maintained constant at 1.0 × 10 –6 M. The OH – from dissolution of Pb(OH)2 is neutralized by the buffer. Therefore we can substitute as follows to obtain the solubility at pH 8. •Calculation [Pb2+ ] = s [Pb2+ ] = Ksp [OH– ]2 1 .2 × 1 0 –15 s= = 1.2 × 10 –3 M (1.0 × 1 0 –6 )2 And at pH = 9, where [OH– ] = 1.0 × 10 –5 M, s= 1.2 × 1 0 –15 = 1.2 × 10 –5 M (1.0 × 1 0 –5 )2 •Comment The lower solubility of Pb(OH) 2 at pH 9.0, than at pH 8.0, is due to the common ion effect. _______________________________________________________________________________ EXAMPLE 16.14 Effect of pH on Solubility Which of the following salts will be more soluble at acidic pH than in pure water? a. AgBr b. Ba(OH) 2 c. MgCO3 •Method of Solution First we need to identify the ions present in solutions of these compounds. Then we identify those with acidbase properties. a. b. AgBr(s) Ag+ + Br – Neither Ag+ nor Br– have acid-base properties and so the solubility of AgBr is not affected by pH. Ba(OH)2 Ba2+ + 2OH – Comparing equilibria in water and in acidic solution, as [H+] increases, [OH– ] decreases, and the solubility equilibrium will shift to the right, causing the solubility of Ba(OH)2 to increase. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 48 / Acid-Base Equilibria and Solubility Equilibria c. Carbonate ion is a base, and in acid solution it combines with H+(aq). 2– Mg 2+ + CO3 MgCO3 2– H+ + CO3 – → HCO3 2– With the increase in [H+] in acid solution, [CO3 ] decreases, and the solubility equilibrium shifts to the right, causing more MgCO3 to dissolve. Answer: (b) and (c) _______________________________________________________________________________ EXERCISES 20. a. What is the molar solubility of Ag3 PO4 in 0.20 M Na3 PO4 ? K sp = 1.8 × 10 –18 b. Write the solubility equilibrium reaction for Ag3 PO4 . 21. What is the molar solubility of Ag2 CrO4 in 0.20 M AgNO3 ? 22. Calculate the molar solubility AgCl in a solution containing 0.010 M CaCl2 . 23. Which of the following salts will be more soluble in an acidic solution than in a basic solution? a. AgI b. BaCO3 c. Ca(CH3 COO)2 d. Zn(OH)2 e. CaCl2 24. What is the solubility of Pb(OH)2 in a solution with a pH of 10.00? Given Ksp = 1.2 × 10 –15 25. Calculate the minimum pH that will just prevent precipitation of Mg(OH)2 from an aqueous solution containing 0.075 M Mg2+ . K sp Mg(OH)2 = 1.2 × 10 –11 COMPLEX IONS AND SOLUBILITY STUDY OBJECTIVES 1. 2. 3. Write the formation constant expression for a given complex ion. Explain the effect of complex ion formation on the solubility of a salt whose metal ion forms a complex ion in solution. Calculate the solubility of a salt in a solution where the metal ion of the salt forms a complex ion. The Formation Constant. When Ag + ion reacts with two CN– ions: – Ag+(aq) + 2CN– (aq) Ag(CN)2 (aq) – it forms a complex ion Ag(CN)2 . An ion made up of the metal ion bonded to one or more molecules or ions is + 2+ 2– called a c omplex ion . Some other examples are Ag(NH3 )2 , Ni(H 2 O)6 , and Cu(CN) 4 . Complex ions are extremely stable and so have an important effect on certain chemical species in solution. A measure of the tendency of a metal ion to form a certain complex ion is given by its formation constant, Kf. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Acid-Base Equilibria and Solubility Equilibria / 3 49 The f ormation constant is the equilibrium constant for the reaction that forms the complex ion. For example, – Ag+ + 2CN– Ag(CN)2 – [Ag(CN)2 ] Kf = = 1.0 × 10 21 [Ag+][CN– ]2 Table 16.4 in the text lists Kf values for selected complex ions. The more stable the complex ion, the greater the extent of reaction, and the greater the value of Kf. The formation of a complex ion has a strong effect on the solubility of a metal salt. For example, AgI, which has a very low solubility in water, will dissolve in an aqueous solution of NaCN. The stepwise process is: Ag+(aq) + I– (aq) AgI(s) – Ag(CN)2 (aq) A g +(aq) + 2CN– (aq) Ksp Kf ___________________________________________________ – overall AgI(s) + 2CN– (aq) Ag(CN)2 (aq) + I– (aq) K The formation of the complex ion in the second step causes a decrease in [Ag+]. Therefore the first equilibrium shifts to the right according to Le Chatelier's principle. This is why AgI is more soluble in CN– solution than in pure water. The overall reaction is the sum of the two steps. The equilibrium constant for the overall reaction is the product of the equilibrium constants of the two steps: K = Ksp × Kf Tables 16.1 and 16.4 of the text give, respectively: Ksp = 8.3 × 10 –17 Kf = 1.0 × 10 21 Therefore K = 8.3 × 10 4 The large value of K shows that AgI is very soluble in NaCN solution. _______________________________________________________________________________ EXAMPLE 16.15 Concentration of Uncomplexed Metal Ion Calculate the concentration of free Ag+ ions in a solution formed by adding 0.20 mol of AgNO3 to 1 L of 1.0 M NaCN. •Method of Solution In this solution Ag+ ions will complex with CN– ions, and the concentration of Ag+ will be determined by the equation Ag+ + 2CN– – Ag(CN)2 K f = 1.0 × 10 21 – Since Kf is so large, we expect the Ag+ to react essentially quantitatively to form 0.20 mol of Ag(CN)2 . To find the concentration of free Ag+ at equilibrium, use the equilibrium constant expression – [Ag(CN)2 ] Kf = [Ag+][CN– ]2 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 50 / Acid-Base Equilibria and Solubility Equilibria Rearranging gives [Ag+] = – [Ag(CN)2 ] Kf[CN– ]2 • Calculation Substitute all known equilibrium concentrations into the above equation. – At equilibrium; [Ag(CN)2 ] = 0.20 M, and – [CN– ] = [CN– ]0 – 2[Ag(CN)2 ] [CN– ] = 1.0 M – 0.40 M = 0.60 M Therefore [Ag+] = (0.20) (1.0 × 1 0 21 )(0.60)2 [Ag+] = 5 × 10 –22 M •Comment This concentration corresponds to only three Ag+ ions per 100 mL! _______________________________________________________________________________ EXAMPLE 16.16 Effect of Complex Formation on Solubility Calculate the molar solubility of silver bromide in 6.0 M NH3 . •Method of Solution The solubility of AgBr is determined by two equilibria. The solubility and complex ion equilibria are AgBr(s) Ag+ + Br – K sp = 7.7 × 10 –13 + A g + + 2NH 3 Ag(NH3 )2 K f = 1.5 × 10 7 ____________________________________________________ + overall AgBr(s) + 2NH3 Ag(NH3 )2 + Br – K = K sp × K f The equilibrium constant of this net reaction controls the solubility of AgBr. K = Ksp × Kf = 1.2 × 10 –5 + K= [Ag(NH3 )2 ][Br– ] [NH3 ]2 • Calculation Let s equal the solubility of AgBr in 6.0 M NH3. _______________________________________________________ + Concentration AgBr(s) + 2NH3 Ag(NH3 )2 + Br – ________________________________________ Initial (M) 6.0 0 0 Change (M) –2s +s +s ________________________________ Equilibrium (M) (6.0 – 2s) s s ______________________________________________________ Substitute into the K expression for the overall reaction Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Acid-Base Equilibria and Solubility Equilibria / 3 51 K= (s)(s) = 1.2 × 10 –5 (6.0 – 2s) 2 The left-hand side is a perfect square. Taking the square root of both sides gives s2 = ( 1.2 × 1 0 –5 ) (6.0 – 2s) 2 s = 3.5 × 10 –3 6.0 – 2s s = 2.1 × 10 –2 M •Comment The solubility of AgBr in pure water is 8.8 × 10 –7 M. The enhanced solubility in this example is due to the + formation of the complex ion Ag(NH3 )2 . _______________________________________________________________________________ EXERCISES 26. Write equations for the formation reaction of each of the following complex ions. 2+ a. Cu(NH3 )4 2– b. Cu(CN) 4 – c. Ag(CN)2 27. Calculate the concentration of Cu2+ ions in a solution formed by adding 0.0500 mol of CuSO4 to 0.500 L of 0.500 M NaCN. 28. Determine the equilibrium constant for the following reaction. – AgBr(s) + 2CN– Ag(CN)2 + Br – 29. Calculate the molar solubility of silver bromide in 1.0 M NH3 . _______________________________________________________________________________ CONCEPTUAL QUESTIONS 1. 2. What is the effect on the pH of the following solutions when: a. solid sodium acetate (CH3 COONa) is added to a dilute solution of acetic acid? b. solid NaCl is added to a dilute solution of NaOH? c. solid KOH is added to a dilute solution of acetic acid? 3. When sodium cyanide (NaCN) is added to a solution of hydrocyanic acid (HCN), the pH: a. increases. b. decreases. c. is unchanged. 4. Back Why is a solution of HBr and NaBr of approximately equal concentrations not considered to be a buffer? Why does an acid-base indicator change color in a titration? Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 52 / Acid-Base Equilibria and Solubility Equilibria 5. The pH at the equivalence point in a titration was found to be approximately 9. What category of titration in terms of acid and base strength is this titration? 6. Explain how changing the pH affects the solubility of Zn(OH) 2 . PRACTICE TEST 1. A buffer solution is prepared by mixing 500 mL of 0.60 M CH3 COOH with 500 mL of 1.00 M CH3 COONa solution. What is the pH of this solution? 2. A buffer solution is prepared by mixing 300 mL of 0.10 M HNO2 with 200 mL of 0.40 M NaNO2 . a. Calculate the pH of the resulting solution. b. What is the new pH after 2.0 mL of 2.0 M HCl are added to this buffer? 3. What is the optimum pH of an HCN / CN– buffer? 4. Which one of the following equimolar mixtures is suitable for making a buffer solution with an optimum pH of about 9.2? a. NaC2 H3 O2 and HC2 H3 O2 b. NH 4 Cl and NH3 c. HF and NaF d. HNO2 and NaNO2 e. NaCl and HCl 5. A 20.0 mL portion of a solution of 0.0200 M HNO3 is titrated with 0.0100 M KOH. a. What is the pH at the equivalence point? b. How many mL of KOH are required to reach the equivalence point? c. What will the pH be after only 10.0 mL of KOH are added? d. What will the pH be after 45.0 mL of KOH are added? 6. Consider the titration of 50.0 mL of 0.10 M CH3 COOH with 0.10 M NaOH. What is the pH at the equivalence point? 7. The color of methyl red indicator changes from red to yellow at pH = 5.2. What is the Ka for the weak acid form of this indicator? 8. The indicator thymol blue is a weak acid with a Ka of approximately 1 × 10 –9 . In 0.1 M HCl the indicator is yellow, and in 0.1 M NaOH it is blue. What color is a solution of pH = 7.0 to which thymol blue has been added? 9. Calculate the pH of a solution prepared by mixing 25.0 mL of 0.10 M HCl and 25.0 mL of 0.25 M CH3 COONa? 10. Which of the following is the least soluble? a. SrF2 Ksp = 2.8 × 10 –9 b. Zn(OH)2 Ksp = 1.8 × 10 –14 c. PbI 2 Ksp = 1.4 × 10 –8 d. BaF2 Ksp = 1.7 × 10 –6 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Acid-Base Equilibria and Solubility Equilibria / 3 53 11. At a certain temperature the solubility of barium chromate (BaCrO4 ) is 1.8 × 10 –5 mol/L. What is the Ksp value at this temperature? 12. If the solubility of Fe(OH) 2 in water is 7.7 × 10 –6 mol/L at a certain temperature, what is its Ksp value at that temperature? 13. What is the molar solubility of silver phosphate (Ag3 PO4 ) in water? Ksp = 1.8 × 10 –18 14. AgCl is most soluble in: a. 0.1 M AgNO 3 b. 0.2 M NaCl c. 2.0 M HCl d. pure water 15. Will a precipitate of BaSO4 form when 400 mL of 0.020 M Na2 SO4 is added to 700 mL of 0.001 M BaCl2 ? K sp (BaSO4 ) = 1.1 × 10 –10 16. Some municipal water supplies contain Ca2+ at a concentration of 3 × 10 –3 M. Will a precipitate of CaSO4 form when 0.10 mol of Na2 SO4 is dissolved in 0.50 L of this water? 17. Will a precipitate of MgF 2 form when 600 mL of solution that is 2.0 × 10 –4 M in MgCl2 is added to 300 mL of 1.1 × 10 –2 M NaF solution? Ksp (MgF 2 ) = 6.4 × 10 –9 18. A solution of NaOH is added dropwise to a solution that is 0.100 M Ca2+ and 0.010 M Mg2+ . a. Which cation precipitates first? b. What concentration of OH– is necessary to begin precipitation? c. What is the concentration of the "first cation to precipitate" when the second cation begins to precipitate? d. What percentage of the initial Mg2+ concentration remains in solution when the Ca2+ begins to precipitate? 19. Which of the following salts should be more soluble in acid solution than in pure water? a. CaCO 3 b . MgCl 2 c. NaNO3 d. LiBr e. KI 20. Which of the following is the most stable complex ion? 3– a. Ag(S 2 O3 )2 Kf = 1.0 × 10 13 + b. Ag(NH3 )2 – c. Ag(CN)2 Kf = 1.5 × 10 7 Kf = 1.0 × 10 21 21. Calculate the concentration of free copper ion in a solution made from 1.0 × 10 –2 M Cu(NO 3 )2 and 1.0 M NH3 . 22. What is the molar solubility of AgCl in 3.0 M NH3 ? 23. What is the molar solubility of AgBr in 0.100 M Na2 S 2 O3 ? Given: 2– Ag+ + 2 S 2 O3 3– Ag(S 2 O3 )2 K f = 1.0 × 10 13 24. a. Will a precipitate form when 400 mL of 0.20 M K2 CrO4 is added to 400 mL of 0.10 M AgNO 3 ? Ksp (Ag2 CrO4 ) = 1.1 × 10 –12 b. If yes, what is the silver ion concentration left in solution? 25. What is the minimum concentration of aqueous NH3 required to prevent AgCl(s) from precipitating from 1.0 L of solution prepared from 0.20 mol of AgNO3 and 0.010 mol of NaCl? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 54 / Acid-Base Equilibria and Solubility Equilibria ANSWERS Exercises 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. Back 1.34% versus 0.09% a. pH = 4.66 b. no 6.50 + a. initial pH is 9.39 and final is 9 45 b. NH4 NH3 + H+ – + NH + → H O + NH c. OH 2 3 4 pH = 2.12 pH > 7. Usually between 8 – 11 depending on the strength of the conjugate base of the weak acid. A titration of a strong acid with a weak base. a. HBr(aq) + NaOH(aq) → H2 O(l) + NaBr(aq) b. pH = 1.40 4.37 yellow pH = 6.8 – – a.BaCO3 (s) Ba2+ (aq) + CO3 (aq) Ksp = [Ba2+ ] [CO3 ] b. CaF2 (s) Ca 2+ (aq) + 2F– (aq) Ksp = [Ca2+ ] [F– ]2 c. Al(OH)3 (s) (aq) + 3OH– (aq) Ksp = [Al3+ ] [F– ]3 3– 3– d. Ag3 PO4 (s) 3Ag+(aq) + PO4 (aq) Ksp = [Ag+]3 [PO 4 ] Ksp = 4.9 × 10 –6 a. s = 2.2 × 10 –4 M b. [F–] = 4.4 × 10 –4 Ksp = 1.7 × 10 –6 No Yes, MgF 2 [Ba2+ ] = 3.2 × 10 –5 M a. CuI(s) b. For CuI to ppt [I– ] must be 5.1 × 10 –8 M. For PbI2 to ppt [I– ] must be 3.7 × 10 –3 M . c. 0.0014% 3– a. s = 2.1 × 10 –6 M b . Ag 3 PO4 (s) 3Ag+ + P O 4 –11 M s = 2.8 × 10 s = 8.0 × 10 –9 M b, c, and d s = 1.2 × 10 –7 M pH = 9.1 2+ a. Cu 2+ (aq) + 4NH3 Cu(NH3 )4 2– b. Cu2+ (aq) + 4CN– Cu(CN)4 – c. Ag +(aq) + 2CN– Ag(CN)2 1.0 × 10 –22 M K = Ksp × K f = 7.7 × 10 8 s = 3.5 × 10 –3 M Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Acid-Base Equilibria and Solubility Equilibria / 3 55 Conceptual Questions 1. 2. 3. 4. A solution of HBr and NaBr cannot be a buffer because the HBr is a strong acid. Its conjugate base Br– ion has no base strength. This means that the buffer cannot neutralize added strong acid. A reaction such as H+ + Br – → HBr cannot occur. a. pH increases b. no change in pH c. pH increases a. increases An indicator is itself a weak acid. The nonionized form HIn has a color we'll call color 1. The conjugate base In– has another color we'll call color 2. When [HIn] > [In – ] the solution is color 1. When [In– ] > [HIn] the solution is color 2. The ionization equilibrium of HIn is: H + + In– HIn 5. 6. In a titration of a strong acid by a strong base, for instance, the H+ ion concentration starts out high so the indicator equilibrium is shifted to the left and the solution is color 1. As the titration continues the solution becomes less acidic and at the equivalence point the pH rises rapidly. As the H+ ion concentration falls dramatically, the indicator equilibrium shifts to the right and its color changes to color 2. weak acid vs. strong base The solubility reaction of Zn(OH)2 is: Zn(OH)2 Zn 2+ (aq) + 2OH– (aq) The pH of a solution directly affects the OH– ion concentration. To adjust the pH upward we must add more OH– ion to a solution. In terms of the solubility equilibrium OH– ion is a common ion and so the solubility equilibrium is shifted to the left as pH increases. Increasing the pH lowers the solubility of Zn(OH)2 . Practice Test 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. pH = 4.96 a. pH = 3.77 b. 3.66 9.31 b. NH 3 Cl and NH3 a. pH = 7.0 b. 40.0 mL c. 2.00 d. 10.89 8.72 Ka = 6.3 × 10 –6 Yellow pH = 4.92 b. Zn(OH)2 Ksp = 3.2 × 10 –10 Ksp = 1.8 × 10 –15 s = 1.6 × 10 –5 M d. pure water Yes Yes No a. Mg 2+ b. [OH– ] = 3.5 × 10 –5 M c. [Mg2+ ] = 1.5 × 10 –7 M a. CaCO 3 – c. Ag(CN)2 2+ ] = 2.35 × 10 –16 M [Cu s = 0.13 M s = 0.042 M a. Yes b. [Ag+] = 3.8 × 10 –6 M [NH3 ] = 0.91 M d. 1.5 × 10 –3 % of Mg 2+ remains. _______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...
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This note was uploaded on 09/15/2009 for the course CHEM 102 taught by Professor Bastos during the Spring '08 term at Adelphi.

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