Question

Consider a 0.20 M solution of the salt, sodium cyanide, NaCN. A) What acid and What base reacted to from this salt? B) Classify the acid above as a strong or weal acid. Classify the base as a strong or weak base? C) Will the salt solution be acidic, basic, or neutural? D) Calculate the ph of the solution? The K_a of HCN is 4.9 x 10^-10?

Answer 1

a) The salt is formed from the strong base NaOH and weak acid HCN

b) NaOH - strong base, HCN - weak acid

c) This salt,NaCN is having basic in nature in its resulting solutions.Because its fromed from strong base.

d) For NaCN (aq) <--------> Na⁺ (aq) + CN^- (aq)

Thus -

           CN^- (aq) + H₂O (l) <-------> HCN (aq) + OH^- (aq)

I(M) 0.20                                         0               0

C        -x                                           +x            +x

Eq     (0.20-x)                                     x              x

Therefore,

                   K_b = (Kw/Ka) = (1.0*10^-14 / 4.9 x 10^-10 ) = (x² / (0.20-x)

by solving - x = [OH^-] = 0.0020 M

Thus , pOH = - log (0.0020) = 2.7

Hence, pH of ths solution = 14 - pOH = 14 - 2.7 = 11.30