11. Propanoic acid (CH3CH2COOH) reacts as a weak acid when dissolved in water, w
ID: 932084 • Letter: 1
Question
11. Propanoic acid (CH3CH2COOH) reacts as a weak acid when dissolved in water, with Ka = 1.7×10^-5 at 25°C. CH3CH2COOH(aq) + H2O() -- CH3CH2CO-(aq) + H3O+ (aq) (4 pts) (a). Write the expression for Ka for propanoic acid in terms of the concentrations of the relevant species. (8 pts) (b). Calculate the percent ionization of propanoic acid in a 0.10 M solution. (4 pts) (c). Suppose we added some 0.10 M HCl to the 0.10 M propanoic acid solution above. What would happen to the concentration of undissociated CH3CH2COOH molecules, and why?
Explanation / Answer
11. Propanoic acid
(a) Ka expression
Ka = [CH3CH2COO-][H3O+]/[CH3CH2COOH]
(b) Calculate [H+] concentration
1.7 x 10^-5 = x^2/0.1
x = [H+] = 1.304 x 10^-3 M
% ionization = ([H+]/[HA]) x 100
= (1.304 x 10^-3/0.1) x 100 = 1.304%
(c) When HCl is added
HCl is a strong acid and dissociated completely
The acidity of solution would be entirely due to addd HCl.
The concentration of CH3CH2COOH would remain the same or negligible dissociation in presence of HCl
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