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Unformatted text preview: Chapter Sixteen ACID-BASE EQUILIBRIA AND SOLUBILITY
• 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
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
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
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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] K
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]
[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
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 =
[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
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–
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
(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
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.
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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
= 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.
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
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:
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
Indicators are usually weak organic acids that have distinctly different colors in the nonionized (molecular)
and ionized forms.
H+(aq) + In– (aq)
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
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
pH = pKa + log (10/1) = pKa + 1 .0
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
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.
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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) +
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.
+ + 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 _________________________________________________
H+ + NH3
(0.050 – x)
Ka = [NH3 ][H+]
[NH4 ] = x2
= 5.6 × 10 –10
(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.
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
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 = [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
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–
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
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
2.4 × 10 –5
1.6 × 10 –10
7.7 × 10 –13
MX2 and M2 X type
2.4 × 10 –4
8.0 × 10 –6
1.4 × 10 –8
6.4 × 10 –9
1.2 × 10 –11
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
s= 7.3 × 1 0 –3 g
100 mL × 1 mL
1 m ol
= 1.17 × 10 –3 mol/L
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
Now substitute the ion concentrations into the Ksp expression.
Ksp = (1.17 × 10 –3 ) (2.34 × 10 –3 )2
Ksp = 6.4 × 10 –9
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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
CrO4 are produced.
[Ag+] = 2s 2– [CrO4 ] = s Summarize the changes in concentrations as follows.
Ag 2 CrO4 (s)
2Ag+(aq) + CrO4 (aq)
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
1 .1 × 1 0 –12
= 2.75 × 1 0 –13
s = 6.5 × 10 –5 M
12. Write balanced equations and solubility product expressions for the solubility equilibria of the following
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
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.
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
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+ ]
Examples 16.10 and 16.11 illustrate the separation of Cl– and CrO4 (chromate) ions.
EXAMPLE 16.9 Predicting Formation of a Precipitate
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 .
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
3 00 mL
[I– ]0 = 0.050 M ×
= 3.0 × 10 –2 M
[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.
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:
1 .4 × 1 0 –8
[Pb2+ ] = – 2 =
= 4.3 × 10 –5 M
The percentage of Pb2+ ion remaining unprecipitated is
4.3 × 10–5 M ( 0.5 L)
× 100% = 0.72%
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
[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 .
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
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%.
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
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 (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
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
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.
PbCl 2 (s)
Pb 2+ (aq) + 2Cl– (aq)
(0.50 + 2s)
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
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.
[Pb2+ ] = s
[Pb2+ ] = Ksp [OH– ]2
1 .2 × 1 0 –15
= 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.
Ag+ + Br – Neither Ag+ nor Br– have acid-base properties and so the solubility of AgBr is
not affected by pH.
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)
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
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
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
– Ag+ + 2CN–
= 1.0 × 10 21
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
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
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
•Method of Solution
In this solution Ag+ ions will complex with CN– ions, and the concentration of Ag+ will be determined by the
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 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 50 / Acid-Base Equilibria and Solubility Equilibria
[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
[Ag+] = (0.20)
(1.0 × 1 0 21 )(0.60)2 [Ag+] = 5 × 10 –22 M
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
Ag+ + Br –
K sp = 7.7 × 10 –13
A g + + 2NH 3
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.
AgBr(s) + 2NH3
Ag(NH3 )2 + Br –
(6.0 – 2s)
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
= 1.2 × 10 –5
(6.0 – 2s) 2 The left-hand side is a perfect square. Taking the square root of both sides gives
= ( 1.2 × 1 0 –5 )
(6.0 – 2s) 2
= 3.5 × 10 –3
6.0 – 2s
s = 2.1 × 10 –2 M
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 .
26. Write equations for the formation reaction of each of the following complex ions.
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
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
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
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
10. Which of the following is the least soluble?
Ksp = 2.8 × 10 –9
Ksp = 1.8 × 10 –14
c. PbI 2
Ksp = 1.4 × 10 –8
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
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
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?
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
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
29. Back 1.34% versus 0.09%
a. pH = 4.66 b. no
a. initial pH is 9.39 and final is 9 45 b. NH4
NH3 + H+
– + NH + → H O + NH
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
pH = 6.8
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
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
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 .
a. s = 2.1 × 10 –6 M b . Ag 3 PO4 (s)
3Ag+ + P O 4
s = 2.8 × 10
s = 8.0 × 10 –9 M
b, c, and d
s = 1.2 × 10 –7 M
pH = 9.1
a. Cu 2+ (aq) + 4NH3
b. Cu2+ (aq) + 4CN–
c. Ag +(aq) + 2CN–
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
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
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
25. pH = 4.96
a. pH = 3.77 b. 3.66
b. NH 3 Cl and NH3
a. pH = 7.0 b. 40.0 mL c. 2.00 d. 10.89
Ka = 6.3 × 10 –6
pH = 4.92
Ksp = 3.2 × 10 –10
Ksp = 1.8 × 10 –15
s = 1.6 × 10 –5 M
d. pure water
a. Mg 2+ b. [OH– ] = 3.5 × 10 –5 M c. [Mg2+ ] = 1.5 × 10 –7 M
a. CaCO 3
2+ ] = 2.35 × 10 –16 M
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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