Question

When 0.1 mol of ammonia $\left(N{H}_3\right)$ is dissolved in sufficient water to make 1 L of solution, the solution is found to have a hydroxide ion $\left(O{H}^{-}\right)$ concentration of $1.5\times {10}^{-3}M$ . The value of $K_b$ for $N{H}_3$ is:

• A. $8.65\times {10}^{-2}M$
• B. $0.5\times {10}^{-3}M$
• C. $2.28\times {10}^{-5}M$
• D. $5.75\times {10}^{-6}M$

Answer 1

The correct option is C $2.28\times {10}^{-5}M$

Concentration of $N{H}_3=\frac{0.1mol}{1L}=0.1mol{L}^{-1}$

$N{H}_3\left(l\right)\stackrel{+H_{2}O}{\rightleftharpoons }N{H}_4^{+}(aq.)+O{H}^{-}(aq.)Initial:0.100Equilibrium:0.1-xxx$
Here, at equilibrium $\left[O{H}^{-}\right]=\left[N{H}_4^{+}\right]=x\therefore \left[N{H}_3\right]=(0.1-x)=0.1-1.5\times {10}^{-3}=0.0985M$

$K_b=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[N{H}_3\right]}=\frac{(1.5\times {10}^{-3}{)}^2}{0.0985}{K}_b=2.28\times {10}^{-5}M$

note: we can also approximate $0.1-x$ as $1$ because $x$ is very less.