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

First, determine the amount of sodium hydroxide present:

$n_{NaOH} = C \times V = 0.222 \, \text{mol/L} \times 0.900 \, \text{L} = 0.1998 \, ext{mol}$

Next, calculate the amount of hydrochloric acid initially added to the flask:

$n_{HCl} = C \times V = 1.00 \, ext{mol/L} \times 0.100 \, ext{L} = 0.1000 \, ext{mol}$

Now, subtract the amount of HCl that reacted with NaOH:

$n_{HCl} ext{ remaining} = 0.1000 - 0.1998 = -0.0998 \, ext{mol}$

Since the amount of NaOH present exceeds that of HCl, all the HCl has reacted, leading to a concentration of:

$C_{HCl} = \frac{n_{HCl}}{1.00 \, ext{L}} = 0.0000 \, ext{mol/L}$

(NOTE: Since the value is theoretically negative, it indicates all HCl was neutralized and NaOH was in excess.)

The calculated concentration of sodium hydrogen carbonate will be greater than the true value. This is because the contaminating NaOH has neutralized all of the HCl, resulting in an insufficient amount of HCl in the solution for the accurate titration of sodium hydrogen carbonate. Thus, the concentration derived from this tainted solution would overestimate the actual concentration.