General Chemistry Math Question 4

Question

Determine the pH at the equivalence point for the titration of 200 mL of 0.0525 M NaCN with 0.0375 M HCl. The Ka for HCN is 4.9 x 10^-10.

Answer 1

NaCN ==> Na⁺(aq) + CN^-(aq)

CN-(aq) + H₂O ==> HCN + OH- since HCN is a weak acid and OH^- is a strong base, this soln will be basic.

Ka x Kb = 1x10^-14 (this is for an acid base conjugate pair)

Kb for CN^- = 1x10^-14 / 4.9x10^-10 = 2.0x10^-5

Kb = 2.0x10^-5 = [HCN][OH-] / [CN-] = (x)(x) / 0.0525 -x (neglect x in denominator assuming it is small)

x² = 1.05x10^-6

x = [OH^-] = 1.02x10^-3 M (assumption was valid as this is only 2% of 0.0525)

1.02x10^-3 mol/L x 0.200 L = 2.04x10^-4 moles OH^-

To neutralize this we need 2.04x10^-4 moles H⁺

(x L HCl)(0.0375 mol HCl/L) = 2.04x10^-4 moles HCl

x = 5.44x10^-3 L = 5.44 mls (ignoring significant figures)