A student is to accurately determine the concentration of a solution of sodium hydrogen carbonate in a titration against a standard solution of hydrochloric acid, HCl - VCE - SSCE Chemistry - Question 3 - 2009 - Paper 1

To find the concentration of hydrochloric acid remaining in the flask:

1. Calculate the moles of NaOH added: $n_{NaOH} = 0.2222 \, \text{M} \times 0.9000 \, \text{L} = 0.20000 \, \text{mol}$

2. Calculate the initial moles of HCl: $n_{HCl} = 1.00 \, \text{M} \times 0.1000 \, \text{L} = 0.10000 \, \text{mol}$

3. Determine the moles of HCl that reacted with NaOH: Since HCl and NaOH react in a 1:1 molar ratio, the amount of HCl that reacted is equal to the moles of NaOH: $n_{HCl \text{ reacted}} = 0.20000 \, \text{mol}$ (the same amount of moles will react, but we only have 0.10000 moles of HCl)

4. Calculate remaining HCl: $n_{HCl \text{ remaining}} = n_{HCl \text{ initial}} - 0.10000 = 0.10000 - 0.20000 = 0.00000 \,(\text{since HCl limits the reaction})$

5. Concentration of HCl in the flask after reaction: $C_{HCl} = \frac{n_{HCl \text{ remaining}}}{V_{total}} = \frac{0.00000}{1.00} = 0.00000 \, \text{M}$

Thus, the concentration of hydrochloric acid after adding sodium hydroxide is 0.00 M.

The calculated concentration of sodium hydrogen carbonate solution will be greater than the true value. This is because the contaminated hydrochloric acid solution has effectively zero concentration of HCl after the NaOH reaction. If the titration uses this solution, it will incorrectly suggest a higher concentration than actually exists, as not enough HCl is present to properly neutralize the sodium hydrogen carbonate.