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 after adding sodium hydroxide, we need to calculate the moles of each substance.

1. Calculate moles of NaOH added: $n(\text{NaOH}) = C \times V = 0.222 \, \text{M} \times 0.900 \, \text{L} = 0.1998 \, \text{mol}$

2. Calculate moles of HCl in the flask:

   • The initial moles of HCl from dilution: $n(\text{HCl, initial}) = 1.00 \, \text{M} \times 0.100 \, \text{L} = 0.100 \, \text{mol}$

3. The reaction will consume HCl and NaOH in a 1:1 ratio:

   • Moles of HCl reacted: $n(\text{HCl reacted}) = n(\text{NaOH}) = 0.1998 \, \text{mol}$

4. Calculate moles of HCl remaining: $n(\text{HCl remaining}) = n(\text{HCl, initial}) - n(\text{HCl reacted}) = 0.100 - 0.1998 = -0.0998 \, \text{mol}$

Since HCl is in excess, we will recalculate the remaining moles based on available moles:

• Thus, HCl is completely reacted, leaving NaOH unreacted.
• The total volume in the flask is 1.00 L.

5. Calculate concentration of HCl in the flask:
   • The concentration of HCl is: $C(\text{HCl}) = \frac{n(\text{HCl remaining})}{\text{total volume}} = \frac{0}{1.00 \, \text{L}} = 0.000 \, \text{M}$
   • This indicates that the HCl in the solution has been neutralized completely, resulting in a negligible concentration.

The calculated concentration of the sodium hydrogen carbonate solution will be greater than the true value. This is because the presence of excess sodium hydroxide in the solution means that the titration may suggest more sodium hydrogen carbonate is present than is actually the case. The NaOH will react with HCl in the titration, leading to a miscalculation of the acid's concentration, which in turn will affect the final determination of the sodium hydrogen carbonate concentration, suggesting it is higher than it should be.