Question

A sample containing ammonium chloride was warmed with 100 mL of 1.00 moldm-3 sodium
hydroxide. After all the ammonia had been driven off, the excess sodium hydroxide required
50.00 mL of 0.250 moldm-3 sulfuric acid for neutralisation. What mass of ammonium chloride did
the sample contain?

Answer 1

When the ammonium chloride reacts with sodium hydroxide, ammonia, sodium chloride and water are formed:

NH₄Cl + NaOH $\rightarrow$ NH₃$\uparrow$ + NaCl + H₂O.

Then, the sulfuric acid reacts with the excess sodium hydroxide to form sodium sulphate and water:

2NaOH + H₂SO₄$\rightarrow$ Na₂SO₄ + 2H₂O.

As one can see, 2 mol of sodium hydroxide react with 1 mol of sulfuric acid:

$\frac{n(NaOH)}{2} = n(H_2SO_4)$ .

The number of the moles of the sulfuric acid can be calculated from its volume and its concentration:

$n(H_2SO_4) = cV = 0.250\text{ mol/dm}^3 \cdot50\cdot10^{-3}\text{ dm}^3$

$n(H_2SO_4) = 0.0125$ mol.

Therefore, the number of the moles of the excess sodium hydroxide was:

$n(NaOH) = 2\cdot0.0125 = 0.025$ mol.

The number of the moles of sodium hydroxide reacted with ammonium chloride can be calculated from the total number of the moles of sodium hydroxide added ($cV = 1.00\text{ mol/dm}^3 \cdot100\cdot10^{-3}\text{ dm}^3 = 0.1$ mol) minus the number of the moles of the excess sodium hydroxide:

$n(NaOH)_{NH_4Cl} = 0.1 - 0.025 = 0.075$ mol.

According to the stoichiometry of the first reaction, the number of the moles of the ammonium chloride is:

$n(NH_4Cl) = n(NaOH) = 0.075$ mol.

Finally, the mass of the sample of the ammonium chloride can be calculated using its molar mass $M$ =53.49 g/mol:

$m(NH_4Cl) = nM = 0.075 \text{ mol}\cdot 53.49 \text{ g/mol}$

$m(NH_4Cl) =4.012 \text{ g}$ .

Answer: the sample contained 4.012 g of ammonium chloride .