Question

Part A) provide the acid dissociation equilibrium reaction for cyanic acid, HOCN
Part B) provide the expression for Kb for the conjugate base of HOCN
Part C) pKa=3.92 for HOCN at 25 degrees . what is the pH of a 1.25M solution of HOCN at this temperature

Answer 1

A) HCNO⇌H⁺+CNO^-

HCN_(aq) + H₂O_(l)  H₃O⁺_(aq) + CN^–_(aq)

B) Solve the equation for Kb by dividing the Kw by the Ka. You then obtain the equation Kb = Kw / Ka. Put the values from the problem into the equation. For example, for the chloride ion, Kb = 1.0 x 10^-14 / 1.0 x 10^6.

C) PH = log 1.25(3.92)

= 0.3798 (10)

= 3.798

PH = 4