Ch18_Buffers - The Common Ion Effect 1 Section 18.2 Section...

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Unformatted text preview: The Common Ion Effect 1 Section 18.2 Section 18.2 After all, NH 4+ iis an acid! After all, NH 4+ s an acid! The Common Ion Effect 3 Section 18.2 QUESTION: What is the effect on the pH of adding NH 4Cl QUESTION: What is the effect on the pH of adding NH 4Cl to 0.25 M NH3((aq)? to 0.25 M NH3 aq)? + NH3(aq) + H2O ¸ NH4 +(aq) + OH--(aq) NH3(aq) (aq) 2O ¸ NH 4 (aq) Let us first calculate the pH of a 0.25 M Let us first calculate the pH of a 0.25 M NH3 solution. NH3 solution. [NH3] [NH4 +] [OH--] [NH3] [NH4 +] [OH ] initial 0.25 0 0 initial 0.25 0 0 change -x +x +x change -x +x +x equilib 0.25 - x x x equilib 0.25 - x x x Here we are adding an ion COMMON to Here we are adding an ion COMMON to the equilibrium. the equilibrium. Le Chatelier predicts that the equilibrium Le Chatelier predicts will shift to the ____________. will shift to the ____________. The pH will go _____________. The pH will go _____________. Section 18.2 2 QUESTION: What is the effect on the pH of adding NH 4Cl QUESTION: What is the effect on the pH of adding NH 4Cl to 0.25 M NH3((aq)? to 0.25 M NH3 aq)? + NH3(aq) + H2O ¸ NH4 +(aq) + OH--(aq) NH3(aq) (aq) 2O ¸ NH 4 (aq) QUESTION: What is the effect on the pH QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH 3(aq )? of adding NH4Cl to 0.25 M NH 3(aq)? + NH3(aq) + H2O ¸ NH4+(aq) + OH--(aq) NH3(aq) (aq) 2O ¸ NH 4 (aq) The Common Ion Effect The Common Ion Effect Kb [NH4+ ][OH- ] x2 = 0.25 - x [NH3 ] 1.8 x 10 -5 = Assuming x is << 0.25, we have [OH-] = x = [Kb(0.25)]1/2 = 0.0021 M This gives pOH = 2.67 and so pH = 14.00 - 2.67 = 11.33 for 0.25 M NH3 11.33 4 The Common Ion Effect Section 18.2 5 The Common Ion Effect Section 18.2 Problem: What is the pH of a solution Problem: What with 0.10 M NH 4Cl and 0.25 M NH3(aq)? with 0.10 M NH 4Cl and 0.25 M NH3(aq)? + NH3(aq) + H2O ¸ NH4+(aq) + OH--(aq) NH3(aq) (aq) 2O ¸ NH4 (aq) Problem: What is the pH of a solution Problem: What with 0.10 M NH 4Cl and 0.25 M NH3(aq)? with 0.10 M NH 4Cl and 0.25 M NH3(aq)? + NH3(aq) + H2O ¸ NH4+(aq) + OH--(aq) NH3(aq) (aq) 2O ¸ NH4 (aq) Problem: What is the pH of a solution with 0.10 M NH4Cl Problem: What is the pH of a solution with 0.10 M NH4Cl and 0.25 M NH3((aq)? and 0.25 M NH3 aq)? + NH3(aq) + H2O ¸ NH4+(aq) + OH--(aq) NH3(aq) (aq) 2O ¸ NH4 (aq) We expect that the pH will decline on We expect that the pH will decline on adding NH 4Cl. Let’s test that! adding NH 4Cl. Let’s test that! [NH3] [NH4 +] [OH--] [NH3] [NH4 +] [OH ] initial initial change change equilib equilib We expect that the pH will decline on We expect that the pH will decline on adding NH 4Cl. Let’s test that! adding NH 4Cl. Let’s test that! [NH3] [NH4 +] [OH--] [NH3] [NH4 +] [OH ] initial 0.25 0.10 0 initial 0.25 0.10 0 change -x +x +x change -x +x +x equilib 0.25 - x 0.10 + x x 0.10 + x x equilib 0.25 - x Kb Page 1 1.8 x 10 -5 = [NH4+ ][OH- ] x (0.10 + x) = 0.25 - x [NH3 ] [OH-] = x = (0.25 / 0.10)K b = 4.5 x 10 -5 M This gives pOH = 4.35 and pH = 9.65 pH drops from 11.33 to 9.65 on adding a common ion. 6 7 8 HCl is added to pure water. 10 Consider HOAc/ OAc- to see how buffers work ACID USES UP ADDED OH We know that OAc- + H2O ¸ HOAc + OHOAc has Kb = 5.6 x 10-10 Therefore, the reverse reaction of the WEAK Therefore, reverse ACID with added OH has Kreverse = 1/ Kb = 1.8 x 109 1.8 Kreverse is VERY LARGE, so HOAc completely snarfs up OH- !!!! 11 Buffer Solutions has Ka = 1.8 x Therefore, the reverse reaction of the WEAK Therefore, reverse BASE with added H + has Kreverse = 1/ Ka = 5.6 x 104 5.6 Kreverse is VERY LARGE, so OAc- completely snarfs up H+ ! Buffer Solutions The function of a buffer is to resist changes in the pH of a solution. Buffers are just a special case of the common ion effect. Buffer Composition Weak Acid + Conj. Base HOAc + OAcH2PO4+ HPO42Weak Base + Conj. Acid NH3 + NH4+ HCl is added to a solution of a weak acid H2PO4- and its conjugate base HPO42-. Consider HOAc/ OAc- to see how buffers work CONJ. BASE USES UP ADDED H + HOAc + H2O ¸ OAc- + H3O+ 9 Buffer Solutions Buffer Solutions Buffer Solutions 12 Buffer Solutions Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [ OAc-] = 0.600 M? HOAc + H2O ¸ OAc- + H3O+ Ka = 1.8 x 10 -5 [HOAc] 10-5 initial change equilib [OAc-] 0.700 -x 0.700 - x 0.600 0 +x +x x 0.600 + x [H3O+] Buffer Solutions Problem: Problem: What is the pH of a buffer that has [ HOAc] = 0.700 M and [OAc-] = 0.600 M? - + H O+ HOAc + H2O ¸ OAc 3 Ka = 1.8 x 10 -5 [HOAc] [OAc-] [H3O+] equilib 0.700 - x 0.600 + x x Assuming that x << 0.700 and 0.600, we have Ka 1.8 x 10 -5 = [H3O + ](0.600) 0.700 [H3O+] = 2.1 x 10-5 and pH = 4.68 pH Page 2 13 14 Buffer Solutions Henderson- Hasselbalch Equation [H3O ] Notice that the expression for calculating the H+ conc. of the buffer is [H3O ] Orig. conc. of HOAc Orig. conc. of OAc- pH [H3O ] [Acid] •K [Conj. base] a [OH ] [Base] •K [Conj. acid] b [Acid] [Conj. base] pH Notice that the H + or OH- concs. depend on K and the ratio of acid and base concs. Adding an Acid to a Buffer pK a - l og pK a + l og [Conj. base] [Acid] The pH is determined largely by the pKa of the pK acid and then adjusted by the ratio of acid and adjusted conjugate base. 16 Adding an Acid to a Buffer pH = 3.00 pH 17 Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (after HCl, pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH before = 4.68) 0.600 (pH What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) 0.600 (pH What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) 0.600 (pH Solution to Part (b) Step 1 — do the stoichiometry H3O+ (from HCl) + OAc- (from buffer) ---> HOAc (from buffer) HOAc The reaction occurs completely because K is very large. Solution to Part (b): Step 1— Stoichiometry [H3O+] + [OAc-] ---> [HOAc] 0.00100 0.600 0.700 Before rxn -0.00100 -0.00100 Change +0.00100 After rxn After rxn 0 0.599 0.701 Solution to Part (b): Step 2—Equilibrium HOAc + H 2O ¸ H3O+ + OAcHOAc [HOAc] ¸ [H3O+] + [OAc-] 0.701 0 0.599 Before rxn -x +x Change +x 0.701-x x 0.599 + x After rxn After rxn Page 3 15 Problem: What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (before HCl, pH = 7.00) b) 1.00 L of buffer that has [ HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) 0.600 (pH Solution to Part (a) Calc. [HCl] after adding 1.00 mL of HCl to 1.00 L of water M1•V1 = M2 • V2 M2 = 1.00 x 10 -3 M [Acid] •K [Conj. base] a Take the negative log of both sides of this Take negative equation • Ka Adding an Acid to a Buffer 18 Adding an Acid to a Buffer 19 Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) -] (pH Solution to Part (b): Step 2—Equilibrium HOAc + H 2O ¸ H3O+ + OAcHOAc [HOAc] [H3O ] After rxn 0.701-x 0.599+x x Because [H 3O+] = 2.1 x 10 -5 M BEFORE adding HCl, we again neglect x relative to 0.701 and we neglect 0.599. [OAc - ] • Ka 0.701 •(1.8 x 10 -5 ) 0.599 You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10 -5 M POSSIBLE ACIDS Ka POSSIBLE HSO4 - / SO421.2 x 10-2 HSO SO HOAc / OAc1.8 x 10-5 HOAc HCN / CN 4.0 x 10-10 HCN Best choice is acetic acid / acetate. [H3O ] 10-5 22 Preparing a Buffer Preparing a Buffer You want to buffer a solution at pH = 4.30. This means [H3O+] = 10-pH = 5.0 x 10 -5 M It is best to choose an acid such that [H 3O+] is about equal to Ka (or pH pKa). —then you get the exact [H 3O+] by adjusting the ratio of acid to conjugate base. = 2.1 x M ------> pH = 4.68 pH The pH has not changed on adding HCl to the buffer! [H3O+] 21 Preparing a Buffer Preparing a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) -] (pH Solution to Part (b): Step 2—Equilibrium HOAc + H 2O ¸ H3O+ + OAcHOAc [HOAc] ¸ [H3O+] + [OAc-] 20 [Acid] •K [Conj. base] a 23 Preparing a Buffer Preparing a Buffer You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10 -5 M [H3O ] 5.0 x 10-5 = [HOAc] [OAc- ] Solve for [ HOAc]/[OAc-] ratio (1.8 x 10 -5 ) 2.78 1 Therefore, if you use 0.100 mol of NaOAc and 0.278 mol of HOAc , you will have pH = 4.30. Page 4 24 Preparing a Buffer Preparing a Buffer A final point — CONCENTRATION of the acid and conjugate base are not important. It is the RATIO OF THE NUMBER OF MOLES It RATIO of each. Result: diluting a buffer solution does Result: diluting not change its pH 25 Preparing a Buffer Preparing a Buffer Buffer prepared from HCO3weak acid CO32conjugate base HCO3- + H 2O ¸ H3O+ + CO32- Page 5 ...
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