Question

70.1 POILSU JUUSTURS USCU (a) Calculate the percent ionization of 0.00820 M hydrocyanic acid (K, - 4.9e-10). % ionization - (b) Calculate the percent ionization of 0.00820 M hydrocyanic acid in a solution containing 0.0380 M sodium cyanide! % ionization -

Answer 1

a)

% ionization = [H⁺] / M * 100%

K_a = [H⁺][CN^-] / [HCN]

4,9 X 10^-10= x* x/ (0.00820 - x)

x= 2 X 10^-6 approximately

so

%ion = [H⁺] / M * 100%

= (2 X 10^-6 / 0.00820) X 100%

%ion =0.0243%

(b)

this is buffer so

pH = pK_a + log(CN^- / HCN)

pK_a= -log(K_a)

= -log(4.9 X 10^-10)

= 9.3098

pH = 9.3098 + log(0.0380 / 0.00820) = 9.976

[H⁺] = 10^-pH = 10^-9.976

= 1.06 X 10^-10

so

% ion = (1.06 X 10^-10) / 0.00820 * 100

% ion= 0.00000128%

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