In a titration of 100 mM hydrocyanic acid (pKa = 9.21) with 100 mM NaOH,?

Question

what will the pH be at the equivalence point?

a. ) 7.0

b. ) 9.0

c. ) 10.0

d. ) 11.0

Answer 1

The answer is d.

pKa = 9.21;  Ka = 6.16 x 10^-10

At the equivalence point, you will have twice the volume of solution and therefore 0.05M NaCN solution. CN- hydrolyses according to the equation:

CN- + H2O ===> HCN + OH-

The hydrolysis constant Kh for this reaction is:

Kh = [HCN][OH-]/[HCN] = Kw/Ka = (1x10^14)(6.16x10^-10) = 1.62 x 10^-5

Let [OH-] = x. Then [HCN] = x, and [CN-] = 0.05 - x. But x is very small compared with 0.05, so we can say that [CN-] = 0.05

x^2/(0.05) = 1.62 x 10^-5, so x^2 = 8.2 x10^-7, and x = 9.01 x 10^-4 = [OH]

[H+] = (1x10^-14)/[OH-], so [H+] = 1.12 x10^-11, and pH = 11.0 to three significant figures.