The ammonium ion content of mixtures can be determined by boiling the mixture with a known excess of sodium hydroxide - HSC - SSCE Chemistry - Question 32 - 2023 - Paper 1

Using the concentration of HCl:

[ n(HCl) = 0.02200 \text{ L} \times 0.1102 \text{ mol L}^{-1} = 2.424 \times 10^{-3} \text{ mol} ]

The moles of NaOH reacting in 20 mL:

[ n(NaOH, \text{in } 20 \text{ mL}) = 2.424 \times 10^{-3} \text{ mol} ]

The total volume in the volumetric flask is 250 mL:

[ n(NaOH, \text{total}) = \frac{250.0}{20.0} \times 2.424 \times 10^{-3} \text{ mol} = 3.030 \times 10^{-2} \text{ mol} ]

Using the initial concentration of NaOH for the 50 mL used:

[ n(NaOH, \text{initial}) = 0.500 \text{ L} \times 1.124 \text{ mol L}^{-1} = 5.620 \times 10^{-1} \text{ mol} ]

The moles of NaOH reacting with ammonium ions:

[ n(NaOH \text{ reacting with } NH_4^+) = n(NaOH, \text{initial}) - n(NaOH, \text{total}) ]

[ = 5.620 \times 10^{-1} - 3.030 \times 10^{-2} = 5.290 \times 10^{-1} \text{ mol} ]

The moles of ammonium ions is equal to the moles of NaOH reacting:

[ n(NH_4^+) = n(NaOH \text{ reacting}) = 5.290 \times 10^{-1} \text{ mol} ]

To find the mass of ammonium ions:

[ m(NH_4^+) = n(NH_4^+) \times 18.042 \text{ g mol}^{-1} ]

[ = 5.290 \times 10^{-1} \times 18.042 = 0.0954 \text{ g} ]