Question

When $0.100$ mole of ammonia, ${NH}_3$, is dissolved in sufficient water to make $1.0L$ of solution, the solution is found to have a hydroxide ion concentration of $1.34\times {10}^{-3}M$. Calculate $K_b$ for ammonia.

Answer 1

$N{H}_3+H_{2}O\rightleftharpoons {NH}_4^{+}+O{H}^{-}$

$\left[N{H}_3\right]=(0.1-1.34\times {10}^{-3})M$

$\left[N{H}_4^{+}\right]=1.34\times {10}^{-3}M$

$\left[O{H}^{-}\right]=1.34\times {10}^{-3}M$

Now

$K_b=\frac{\left[{NH}_4^{+}\right]\left[O{H}^{-}\right]}{\left[{NH}_3\right]}$

$K_b=1.82\times {10}^{-5}$