Question

calculate the mass of NaCN that must be added to 250.0 mL of water in order to obtain a solution having a pH of 11.25? Also, calculate the % ionization of the solution. [Ka of HCN=4.9*10^-10]

Answer 1

NACN = Na+ and CN-

CN- + H2O <-> HCN + OH-

Kb = [HCN ][OH-]/[CN-]

Kb = Kw/Ka = (10^-14)/(4.9*10^-10) =0.00002040816 = 2.04*10^-5

pH = 11.25

pOH = 14-11.25 = 2.75

[OH-] = 10^-pOH = 10^-2.75 = 0.00177827

then

Kb = [HCN ][OH-]/[CN-]

2.04*10^-5 =(0.00177827)(0.00177827)/(M - 0.00177827)

M = (0.00177827^2)/(2.04*10^-5) + 0.00177827

M = 0.15679024024 M of NACN

mol = MV = 0.15679024024 *0.25 = 0.039197 mol

mass = mol*MW = 0.039197*49.0 = 1.920653 g of NaCN needed

then

%ion = OH-/M * 100 = 0.00177827/0.1567902*100 = 1.1341 %