The solubility of highly soluble, thermally unstable salts such as ammonium chloride may be determined by back titration - VCE - SSCE Chemistry - Question 8 - 2012 - Paper 1

The amount of NaOH can be calculated using the formula:

$n = C \times V$

Where:

• $C = 0.400\, \text{mol L}^{-1}$ (concentration of NaOH)
• $V = 0.0100\, \text{L}$ (volume of NaOH solution)

Substituting the values:

$n_{NaOH} = 0.400\, \text{mol L}^{-1} \times 0.0100\, \text{L} = 4.00 \times 10^{-3}\, \text{mol}$

Using the equation, the amount of ammonium chloride can be determined. After the reaction with NaOH, the excess hydroxide was neutralised by HCl:

First, calculate the moles of HCl used:

$n_{HCl} = C \times V = 0.125\, \text{mol L}^{-1} \times 0.0147\, \text{L} = 1.84 \times 10^{-3}\, \text{mol}$

Next, calculate moles of NaOH in excess after reaction with NH4Cl:

$n_{NaOH~in~excess} = n_{NaOH} - n_{HCl} = (4.00 \times 10^{-3}) - (1.84 \times 10^{-3}) = 2.16 \times 10^{-3}\, \text{mol}$

Finally, equate this to the ammonium chloride neutralised by NaOH:

$n_{NH4Cl} = n_{HCl} = 1.84 \times 10^{-3}\, \text{mol}$

To find the amount of ammonium chloride in 5.00 mL saturated solution, we first need the total moles in the 20.0 mL aliquot:

$n_{NH4Cl~in~20.0~mL} = 1.84 \times 10^{-3}\, \text{mol}$

Using the dilution factor:

$\text{Dilution~Factor} = \frac{250.0\, \text{mL}}{20.0\, \text{mL}} = 12.5$

Thus, $n_{NH4Cl~in~250.0~mL} = 1.84 \times 10^{-3}\, \text{mol} \times 12.5 = 2.30 \times 10^{-2}\, \text{mol}$

For 5.00 mL:

$n_{NH4Cl~in~5.00~mL} = \frac{2.30 \times 10^{-2}\, \text{mol}}{250.0~mL} \times 5.00~mL = 4.60 \times 10^{-4}\,\text{mol}$

To find the solubility, we first calculate the mass of ammonium chloride:

$\text{Mass} = n \times \text{Molar Mass} = 2.30 \times 10^{-2}\, \text{mol} \times 53.5\, \text{g mol}^{-1} = 1.23\,\text{g}$

Now calculate solubility in gL−1:

$\text{Solubility} = \frac{1.23\,\text{g}}{0.250\,\text{L}} = 4.92\,\text{g L}^{-1}$

Rinsing the burette with water could dilute the hydrochloric acid used for neutralisation. This would result in a lower concentration of HCl reaching the endpoint of the titration. As a result, the calculated amount of NaOH would appear to be higher, implying that more NH4Cl was present in the solution than actually was. Therefore, the calculated solubility of ammonium chloride would be inaccurately high, as the total amount of solid dissolved would have been misrepresented by the dilution effects.