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# More work with Weak Acids, Ka, and pH April 25, 2012.

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More work with Weak Acids, Ka, and pH April 25, 2012

October 11, 20142 Second Niacin problem The K a for niacin is 1.6 x 10 -5. What is the pH of a 0.010 M solution of niacin? 1 st find the [H + ] at equilibrium NiacinH+H+ niacin ion Initial0.01000 Change-x+x Equilibrium 0.010-xxx

October 11, 20143 K a = [H + ] [niacin ion] = 1.6 x 10 -5 [niacin] 1.6 x 10 -5 = x 2 / (0.010-x) x 2 + 1.6 x 10 -5 x - 1.6 x 10 -7 = 0 x = 3.92 x 10 -4 = [H + ] pH = -log(3.92 x 10 –4 ) pH = 3.41

October 11, 20144 Typical behavior of weak acids The concentration of H + is only a small fraction of the concentration of the acid in solution. Relates directly to acid properties, like – Electrical conductivity – Rate of reaction with an active metal

October 11, 20145 The percent ionization of a weak acid _____________ as it concentration increases.

October 11, 20146 Check this out to see how the numbers work. Calculate the percentage of HF molecules ionized in – A. a 0.10 M HF solution – B. a 0.010 M HF solution HF (aq)  H + (aq) + F - (aq) K a = 6.8 x 10 -4

October 11, 20147 HF (aq)  H + (aq) + F - (aq) I 0.10 M 0 M 0 M C - x + x + x E 0.10 – x x x Solve x 2 = 6.8 x 10 -4 (0.10 –x) X = 7.9 x 10 -3 M [H + ]

October 11, 20148 Now, for percent ionization = conc ionized x 100% original conc = 7.9 x 10 -3 M x 100% = 7.9% 0.10 M Repeat the steps for concentration of 0.010 M: X = 2.3 x 10 -3 M % ionization = 23 %

October 11, 20149 How can this be? It is what we would expect from Le Chatelier’s Principle. HF (aq)  H + (aq) + F - (aq) Dilution causes the concentration of the products to decrease, To make up for that decrease, the equilibrium shifts to the right (product side). Increases the amount of ionization.

A 0.0500 M weak acid solution, HQ is only 4.5% ionized. Calculate the equilibrium [H + ], [Q - ], [HQ], pH, and Ka for the acid HQ. HQ  H + + Q - October 11, 201410 [H + ]=[Q - ] = 0.0025 M Q - [HQ] at equilibrium = 0.0500 – 0.0025 = 0.0478 M

Continued…. October 11, 201411 pH = -log(0.0025) = 2.6

October 11, 201412 The solubility of CO 2 in pure water at 25 0 c and 1 atm pressure is 0.0037 M. The common practice is to assume that all the of the dissolved CO 2 is in the form of carbonic acid (H 2 CO 3 ). CO 2 (aq) + H 2 O  H 2 CO 3 (aq) What is the pH of a 0.0037 M solution of H 2 CO 3 ?

October 11, 201413 H 2 CO 3 is a diprotic acid, and K a1 (4.3 x 10 -7 ) and K a2 (5.6 x 10 -11 ) differ by more than a factor of 10 3. The pH can be determined by considering only K a1. H 2 CO 3 H+H+ HCO 3 1- Initial0.0037 M00 Change-x+x Equilibrium0.0037-xxx

October 11, 201414 K a1 = 4.3 x 10 -7 = [H + ][HCO 3 1- ] [H 2 CO 3 ] 4.3 x 10 -7 = x 2 / 0.0037 X 2 = (4.3 x 10 -7 )(0.0037) x = 3.99 x 10 -5 = [H + ] pH = -log (3.99 x 10 -5 ) = 4.40

Homework Chapter 16 Problems 43, 44, 46, 48, 50, 52, 54

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