Question

A 50.00 mL sample of an ammonia solution is analyzed by titration with HCI. The reaction is

${\mathbf{NH}}_3\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{H}}^{+}\left(\mathrm{aq}\right)\to {{\mathbf{NH}}_4}^{+}\left(\mathrm{aq}\right)$

It took 39.47 mL of 0.0984 M HCI to titrate (react completely with) the ammonia. What is the concentration of the original ammonia solution?

Short answer

The concentration of the original ammonia solution is 0.077 M.

Step-by-step solution

Definition of ammonia solution

Ammonia is used as arefrigerant gas, for purification of water supplies, and also in the manufacturing of plastics, explosives, textiles, pesticides, dyes, and other chemicals.

Calculation of moles ofHCI

$N{H}_3\left(aq\right)+H^{+}\left(aq\right)\to N{H_4}^{+}\left(aq\right)$

$\begin{array}{l}\text{Molesof}HCI=\frac{0.0984molHCI}{1.0LHCI}\times \frac{1.0L}{1000mL}\times 39.47mLHCI\\ \text{Molesof}HCI=3.88\times {10}^{-3}molHCI\end{array}$

Hence, the number of molesis$3.88\times {10}^{-3}molHCI$ .

Calculation of molesof NH3

$\begin{array}{l}\text{Molarityof}N{H}_3=\frac{3.88\times {10}^{-3}molHCI}{50.0mLN{H}_3}\times \frac{1.00molN{H}_3}{1.00molHCI}\times \frac{1000mLN{H}_3}{1.00LN{H}_3}\\ \text{Molarityof}N{H}_3=0.077M\end{array}$

Hence, the number of moles is 0.077M.