Find the concentration of [OH-] in the following bases:

Question

a) The [OH^-] of a solution of 0.078 mol/L ammonia is a.bcde mol/L

b) The [OH^-] of a solution of 3.99 mol/L sodium carbonate is a.bcd mol/L

Answer 1

a) The Kb for NH3 is 1.8x10^-5 (Wikipedia)

NH3 + H2O ==> NH4+ + OH-

Kb = [NH4+][OH-] / [NH3] = (x)(x) / 0.078 - x = 1.8x10^-5 (assume x is small and ignore it in denominator)

x² = 1.404x10^-6 (this is only about 1.5% of the 0.078 so assumption above was valid)

x = 1.185x10^-3 mol/L = [OH-]

b) Na2CO3 + 2H2O ==> 2NaOH + H2O + CO2

3.99 mol/L Na2CO3 = 2x3.99 mol/L Na+ and 2x3.99 mol/L OH- = 7.98 mol/L OH-