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More work with Weak Acids, Ka, and pH

More work with Weak Acids, Ka, and pH. April 25, 2012. 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. K a = [H + ] [niacin ion] = 1.6 x 10 -5 [niacin]

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More work with Weak Acids, Ka, and pH

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

  2. Second Niacin problem The Ka for niacin is 1.6 x 10-5. What is the pH of a 0.010 M solution of niacin? 1st find the [H+] at equilibrium

  3. Ka = [H+] [niacin ion] = 1.6 x 10-5 [niacin] 1.6 x 10-5 = x2 / (0.010-x) x2 + 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

  4. 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

  5. The percent ionization of a weak acid _____________ as it concentration increases.

  6. 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) • Ka = 6.8 x 10-4

  7. 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 x2 = 6.8 x 10-4 (0.10 –x) X = 7.9 x 10-3M [H+]

  8. 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 10-3M % ionization = 23 %

  9. 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.

  10. 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- [H+]=[Q-] = 0.0025 M Q- [HQ] at equilibrium = 0.0500 – 0.0025 = 0.0478 M

  11. Continued…. • pH = -log(0.0025) = 2.6

  12. The solubility of CO2 in pure water at 250c and 1 atm pressure is 0.0037 M. The common practice is to assume that all the of the dissolved CO2 is in the form of carbonic acid (H2CO3). • CO2 (aq) + H2O  H2CO3 (aq) • What is the pH of a 0.0037 M solution of H2CO3?

  13. H2CO3 is a diprotic acid, and Ka1 (4.3 x 10-7) and Ka2 (5.6 x 10-11) differ by more than a factor of 103. The pH can be determined by considering only Ka1.

  14. Ka1 = 4.3 x 10-7 = [H+][HCO31-] [H2CO3] • 4.3 x 10-7 = x2 / 0.0037 • X2 = (4.3 x 10-7)(0.0037) • x = 3.99 x 10-5 = [H+] • pH = -log (3.99 x 10-5 ) = 4.40

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

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