Question

1. (a). What are the pH and the percent ionization of an aqueous 0.100 M methanoic acid, HC(O)OH(aq), solution? Explain your reasoning. Note: Ka = 1.8 x 10-4 for methanoic acid. (b). Predict, using Le Châtelier's Principle, whether the percent ionization of 0.050 M methanoic acid will be larger, smaller, or the same as for the 0.100 M solution in part (a). Explain your reasoning. (c). Calculate the percent ionization for the 0.050 M methanoic acid solution. Was your prediction in part (b) correct? Explain why or why not.

Answer 1

1.

a)

Ka = 1.8 x 10^-4

concentration = 0.100 M

[H+] = sqrt (Ka x C)

       = sqrt (1.8 x 10^-4 x 0.1)

       = 4.24 x 10^-3 M

pH = -log [H+] = -log (4.24 x 10^-3)

     = 2.37

pH = 2.37

% ionization = [H+] / C x 100

                    = 4.24 x 10^-3 / 0.1 x 100

% ionization = 4.24 %

b)

HCOOH ------------------> HCOO-   + H+

0.05                                     0            0

0.05 - x                                  x             x

Ka = x^2 / 0.05 - x

1.8 x 10^-4 = x^2 / 0.05 - x

x = 2.91 x 10^-3

% ionization = x / C x 100

                   = 2.91 x 10^-3 / 0.05 x 100

                    = 5.82 %

no percent ionization of 0.1 M methanoic acid is not same as 0.05 M methanoic acid