This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Feb. 25, 2008 Lecture 16, Chapter 17: Further Equilibria Typical Exam Question, 3 parts: (a) pH of 0.30 M H(OAc) soln if K a = 1.8x10-5 (b) pH of 0.30 M Na(OAc) soln (c) pH of 0.30 M H(OAc) and 0.30 M Na(OAc) in same container H(OAc) (aq) H + (aq) + (OAc)- (aq) K a = [H + ][OAc- ] [H(OAc)] (a) I 0.30 M 0 M 0 M C- x + x + x E (0.30 x) (x) (x) (x) 2 (0.30-x) = (x) 2 (0.30) = (0.30)(1.8x10 5 ) (x 2 K a = [H + ][OAc- ] [H(OAc)] If x small = 1.8x10-5 pH = -log (2.3x10-3 ) = (0.30)(1.8x10-5 ) (x) = 2.3x10-3 = [H+] x pH = 2.64 (acidic) (OAc)- (aq) OH- (aq) + H(OAc) (aq) K b = [OH- ][H(OAc)] [(OAc)-] (b) I 0.30 M 0 M 0 M C- x + x + x E (0.30 x) (x) (x) (x) 2 (0.30-x) = (x) 2 (0.30) = (0.30)(5.6x10-10 ) (x) 2 If x small = 5.6x10-10 = 1.3x10-5 = [OH- ] x pOH = -log (1.3x10-5 ) pOH = 4.90 K b = K w K a 1.0x10-14 1.8x10-5 K b = 5.6x10-10 K b = K b = [OH- ][H(OAc)] [(OAc)-] pH = 9.10 (basic) pH + pOH = 14 pH = 14- pOH = 14 - 4.90 By mixing the two solutions it seems like we should probably end up with an acidic solution...
View Full Document
This note was uploaded on 04/17/2008 for the course CHEM 2 taught by Professor Ryan during the Spring '08 term at Marquette.
- Spring '08