Help with ph and percent ionization?

0.035 mol of solid C3H7COONa is added to 450 mL of 0.10 M C3H7COOH. Assume no change in the volume of the solution. Ka for C3H7COOH is 1.52*10-5



a. How much will the percent ionization of the butanoic acid change? Give the absolute value.



b. How much will the pH change?Help with ph and percent ionization?
In the solution of butanoic acid;



C3H7COOH(a) %26lt;------%26gt; C3H7COO-(aq) + H+(aq)

0.10 M - x ................................ x M ............... x M



Ka = [C3H7COO-][H+] / [C3H7COOH] = 1.52x10^-5



(x)(x) / (0.10 - x) = 1.52x10^-5



Since x is very small compared to 0.10, x can be neglected.

x^2 = 1.52 x10^-6

x = 1.23x10^-3 M = [H+]



pH = -log[H+] = -log(1.23x10^-3) = 2.91



% ionization = (1.23x10^-3 / 0.1)x100 = 1.23 %



When sodium butanoate (C3H7COONa) is added to the acid solution, it dissociates completely;

C3H7COONa(aq) -----%26gt; C3H7COO-(aq) + Na+(aq)

0.035 mol .......................... 0.035 mol



Molarity of C3H7COO- ion = n / V = 0.035 mol / 0.450 L

= 0.078 M

The concentration of C3H7COO- ion (0.078 M) produced by the dissociation of the salt is much more larger than that of produced by the dissociation of the acid (1.23x10^-3). Therefore we can neglect 1.23x10^-3 M (The common ion effect).



Ka = [C3H7COO-][H+] / [C3H7COOH] = 1.52x10^-5

(0.078)[H+] / (0.10) = 1.52x10^-5



[H+] = 1.95x10^-5 M



pH = 4.71



pH increases from 2.91 to 4.71



% ionization = (1.95x10^-5 / 0.1)x100 = 0.02 %



Percent ionization decreases from 1.23% to 0.02%Help with ph and percent ionization?
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