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

Jun 20, 2016

$\left[N {H}_{4}^{+}\right] = 3.54 \times {10}^{- 5} \text{ ""mol/l}$

#### Explanation:

Ammonia is a weak base and partially ionises in water:

$N {H}_{3} + {H}_{2} O r i g h t \le f t h a r p \infty n s N {H}_{4}^{+} + O {H}^{-}$

For which:

${K}_{b} = \frac{\left[N {H}_{4}^{+}\right] \left[O {H}^{-}\right]}{\left[N {H}_{3}\right]} = 1.77 \times {10}^{- 5} \text{ ""mol/l}$

$\therefore \left[N {H}_{4}^{+}\right] = {K}_{b} \times \frac{\left[N {H}_{3}\right]}{\left[O {H}^{-}\right]}$

The concentrations given are at equilibrium so:

$\left[N {H}_{4}^{+}\right] = 1.77 \times {10}^{- 5} \times \frac{0.02}{0.01} = 3.54 \times {10}^{- 5} \text{ ""mol/l}$