Question

What is the original molarity of a solution of ammonia whose pH is 10.46?

Answer 1

ammonia = NH3(aq)

in soution :

NH3(aq) + H2O(l) <-> NH4+(aq) + OH-

there are free OH- ions so, expect a basic pH

B + H2O <-> HB+ + OH-

The equilibrium Kb:

Kb = [HB+][OH-]/[B]

Kb = [NH4+][OH-]/[NH3]

initially:

[HB+] = 0

[OH-] = 0

[B] = M

the change

[HB+] = x

[OH-] = x

[B] = - x

in equilibrium

[HB+] = 0 + x

[OH-] = 0 + x

[B] = M - x

we know pH

pH = 10.46

pOH = 14-pH = 14-10.46 = 3.54

[OH-] = 10^-pOH =10^-3.54 = 0.0002884

so

x = 0.0002884

[HB+] = 0 + x = 0.0002884

[OH-] = 0 + x = 0.0002884

[B] = M - x = M- 0.0002884

Kb for NH3 = 1.8*10^-5

substitute

Kb = [NH4+][OH-]/[NH3]

1.8*10^-5 = 0.0002884*0.0002884/(M-0.0002884)

M = (0.0002884^2)/( 1.8*10^-5) + 0.0002884

M = 0.00490 for NH3