Question

How many moles of hydroxide will be formed if 0.500 g of urea are hydrolyzed and the final pH of the solution is 9.25, the pKa of ammonium (NH4+)

Answer 1

pKa of NH4+ = 9.25

pKb = 14 - pH = 4.75

pKb = -log[Kb]

Kb = 1.8 x 10^-5

Hydrolysis of urea,

(NH2)2CO + H2O ---> 2NH3 + CO2

moles of urea = 0.500 g/60.06 g/mol = 0.0083 mols

let us say we have 1 L of solution, so,

[urea] present = 0.0083 mols/1 L = 0.0083 M

NH3 + H2O <==> NH4+ + OH-

Kb = x^2/0.0083

1.8 x 10^-5 = x^2/0.0083

x = [NH4+] = [OH-] = 3.86 x 10^-4 M

moles of OH- formed = 3.86 x 10^-4 M x 1 L = 3.86 x 10^-4 mols

pOH = -log[OH-] = 3.413

pH = 14 - pOH = 10.587