acid–base titrations

Dilute sulfuric acid and aqueous sodium hydroxide are used to make aqueous sodium sulfate，$$\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})$$，or aqueous sodium hydrogen sulfate，$$\mathrm{NaHSO}_{4}(\mathrm{aq})$$．The method includes use of the following apparatus． $$25.0 \mathrm{~cm}^{3}$$ of aqueous sodium hydroxide of concentration $$0.100 \mathrm{~mol} / \mathrm{dm}^{3}$$ was neutralised by $$25.0 \mathrm{~cm}^{3}$$ of dilute sulfuric acid of concentration $$0.0500 \mathrm{~mol} / \mathrm{dm}^{3}$$．The equation for the reaction is shown．This is reaction $$1 .$$ $$ 2 \mathrm{NaOH}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\quad\quad reaction 1 $$ The same technique and the same solutions can be used to make aqueous sodium hydrogen sulfate．The equation for the reaction is shown．This is reaction $$2 .$$ $$ \mathrm{NaOH}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow \mathrm{NaHSO}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I})\quad\quad reaction 2 $$ Complete the table to calculate the volume of dilute sulfuric acid that reacts with $$25.0 \mathrm{~cm}^{3}$$ of aqueous sodium hydroxide in reaction 2． $$ \begin{array}{|l|c|c|} \hline & \begin{array}{c} \text { volume of } 0.0500 \mathrm{~mol} / \mathrm{dm}^{3} \\ \text { dilute sulfuric acid in } \mathrm{cm}^{3} \end{array} & \begin{array}{c} \text { volume of } 0.100 \mathrm{~mol} / \mathrm{dm}^{3} \\ \text { aqueous sodium hydroxide in } \mathrm{cm}^{3} \end{array} \\ \hline \text { reaction 1 } & 25.0 & 25.0 \\ \hline \text { reaction 2 } & & 25.0 \\ \hline \end{array} $$