1. (a). What are the pH and the percent ionization of an aqueous 0.100 M methano
ID: 971869 • Letter: 1
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.
Explanation / Answer
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
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