I need help with these chemistry problems

Question

1. The Ka for cyanic acid (HCN) is 6.2 x 10-10. Determine the pH of a 0.025 M NH4CN salt solution.
2. Which of the following expressions correctly gives the Kb for the fluoride ion, based on the equation given below?

HF (aq) ↔ H+ (aq) + F- (aq) Ka = 7.2 x 10-4

(A) (1.00 x 10-14) x (7.2 x 10-4)

(B) (7.2 x 10-4) / (1.00 x 10-14)

(C) (1.00 x 10-14) / (7.2 x 10-14)

(D) The Kb for the fluoride ion cannot be found from this information

Answer 1

1). NH₄CN is the salt of a strong base (NH₄OH) and a weak acid (HCN). The pH of a solution will therefore be alkaline (pH > 7). One must look at the hydrolysis of the CN^- ion, as follows;

CN^- + H₂O ==> HCN + OH^-

Ka HCN = 6.2x10^-10, so the Kb CN^- = 1x10^-14 / 6.2x10^-10 = 1.6x10^-5

Kb = 1.6x10^-5 = [HCN][OH^-]/[CN^-] = (x)(x) / (0.025 - x) assuming x is small we can ignore it in the denominator

x² = 4x10^-7

x = 6.3x10^-4 M (which is small compared to 0.025 M, so assumption was valid)

[OH-] = 6.3x10-4 M

pOH = -log 6.3x10-4 = 3.2

pH = 14 - pOH

pH = 10.8

2). The F^- ion is the conjugate base of the acid HF

From the expression Kw = KaKb, we know that Kb = Kw/Ka and Kw = 1x10^-14 (ionization of water). We also know Ka = 7.2x10^-4, so the correct expression for the Kb of F^- = 1x10^-14 / 7.2x10^-4 (option A)