Answer to Question #141668 in General Chemistry for sarah

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