# cehmhelp - Chapter 16 Acid – Base Equilibria...

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Unformatted text preview: Chapter 16 Acid – Base Equilibria End-of-Chapter Problems: See Next Page; will discuss along with Ch 15 when through with Ch 16 ~ Oct 9 Online Assignment #1, Chaps 15 & 16; Due: Th Oct 15 11:30 PM Exam #1 over 15 & 16, Lab #1 & Weak Acid Labs: Mon Oct 19 End-of-Chapter Problems pp 690 - 698, Chapter 16 Finish these on or before the last ch. 16 lecture 1, 2, 3, 5, 6; 8 through 20; 23, 24, 26, 29, 30, 31, 32, 35, 37, 41, 43, 50, 59, 60, 63, 65, 71, 75, 79, 85, 87, 115, 117a (& Ka if pH=4.89 - half way to end pt), 120 I. Ka Values A. Introduction- In the preceding section we calculated the [H + ] if a strong acid (or base) was added to water. The major source of [H + ] was from the 100 % ionization of the strong acid.- Now we will look at weak acids (HA) which only partially ionize. In order to determine the [H + ], we will need to use the equilibrium expression.- HA H + + A- or HA (-----) H + + A-- Ka = [H + ] [A- ] [HA] I. Ka Values A. Introduction & B. Obtaining Ka = Large for Strong Acids, > 1 Ka = Small for Weak Acids, < 1 Examples: (only need to memorize Kw for water) + H + H H O- H H H H H OH H H H HA A- + H + 0.10 M Phenol (HA); pH = 5.43; Determine: a ) Ka & b ) % Ionization--3 II. Calculations Involving HA Introduction - two solution methods – 1) quadratic & 2) approximation- Once Ka is known, then can calculate equilibrium concentrations of HA, A- , H +- Will frequently let X = mole/L ( M ) of HA that dissociates- May end up with quadratic equation when solving for X & solve two ways: 1) Quadratic: for a x 2 + b x + c = 2) Approximation: Simplify math if causes less than a 5% error. Can check approximation by either of two methods: a) make approx. & check to see if x is less than 5% of what comparing it to, or use book method : b) if [HA] / Ka > 100 , then approximation OK; this last method allows you to check before doing math. a ac b b x 2 / )] 4 ( [ 2- ±- = . Write equil. rxn & equil. expression with given chemical as reactant. If have to reverse, then : K a = 10-14 / K b . If a rxn takes place, then allow to occur & calculate the new initial II. Calculations Involving HA – Outline of rest of chapter Note: All problems worked by method on previous page A. Weak Acids B. Weak Bases C. Salts of Weak Acids & Salts of Weak Bases D. Common Ion (weak acid or base plus the salt) E. Buffers (same as common ion problem) F. Titrations (1. allow rxn to occur & calc. new initial M ) (2. determine equilibria values using one of A. Overview: Weak Acid (only HA) HA H + + A- Ka = Small- Given Y = init. M of HA; let x = M of HA that ionizes- Drop x if small wrt to Y - if [HA] / Ka > 100 Ka = [H + ][A- ] = x 2 ≈ x 2 [HA] [Y-x] [Y] x = √ [Y] Ka A. Weak Acids - Example A. Weak Acids - Notes Notes: 1) For 0.10 M acetic acid with a Ka = 1.7x10-5 , the approximation was OK....
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cehmhelp - Chapter 16 Acid – Base Equilibria...

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