Question

What is `[NH_4^+]` in a solution that is 0.0200 M `NH_3` and .0100 M `KOH`?

Answer 1

`[NH_4^+]=3.54xx10^(-5)" ""mol/l"`

Explanation:

Ammonia is a weak base and partially ionises in water:

`NH_3+H_2OrightleftharpoonsNH_4^++OH^-`

For which:

`K_b=([NH_4^+][OH^-])/([NH_3])=1.77xx10^(-5)" ""mol/l"`

`:.[NH_4^+]=K_bxx[[NH_3]]/[[OH^-]]`

The concentrations given are at equilibrium so:

`[NH_4^+]=1.77xx10^(-5)xx0.02/0.01=3.54xx10^(-5)" ""mol/l"`