Answer
$V(H_{2}SO_{4}) =21.8ml$
$V(H_{2}O) = 69.9ml$
Work Step by Step
The following reaction occurs:
$H_{2}SO_{4}+2NaOH \rightarrow Na_{2}SO4 + 2H_{2}O$
We can conclude that the amount of sulfuric acid necessary to react with sodium - hydroxide is two times less than the amount of $NaOH$, considering stoichiometric coefficients.
$n(H_{2}SO_{4}) = \frac{1}{2}n(NaOH) = \frac{1}{2}c(NaOH)V(NaOH)=\frac{1}{2}\times6\frac{mol}{dm^3}\times0.0655dm^3=0.1965mol$
We can obtain the volume of the $H_{2}SO_{4}$ solution knowing its concentration:
$V(H_{2}SO_{4}) =\frac{n(H_{2}SO_{4})}{c(H_{2}SO_{4})} = \frac{0.1965mol}{9\frac{mol}{dm^3}} = 0.0218dm^3=21.8ml$
Amount of sodium - sulfate obtained by the reaction above equals to the amount of sulfuric reaction that has reacted:
$n(Na_{2}SO_{4}) = 0.1965mol$
If the concentration of our solution needs to be $1.25M$, its volume needs to be $V = \frac{n}{c} = \frac{0.1965mol}{1.25\frac{mol}{dm^3}}=0.1572dm^3=157.2ml$
Supposing that the volumes are additive, the final volume of the solution obtained by the reaction equals $V(H_{2}SO_{4}) + V(NaOH) = 21.8ml + 65.5ml = 87.3ml$. Therefore, the volume of the water that needs to be added equals $157.2 ml - 87.3ml = 69.9ml$
