Answer
$c(H_{2}SO_{4})=2.28\frac{mol}{dm^3}$
$V(H_{2}SO_{4}) = 1.15ml$
Work Step by Step
$\omega=19.6\%$
$\rho=1.14\frac{g}{cm^3}$---$c=?$
$c(H_{2}SO_{4})=\frac{n(H_{2}SO_{4})}{V(H_{2}SO_{4})}=\frac{m(H_{2}SO_{4})}{M(H_{2}SO_{4})\times V(H_{2}SO_{4})}=\frac{\omega \times m_{sol}}{M(H_{2}SO_{4})\times V(H_{2}SO_{4})}=\frac{\omega \times \rho}{M(H_{2}SO_{4})} = \frac{0.196 \times 1140 \frac{g}{dm^3}}{98\frac{g}{mol}}=2.28\frac{mol}{dm^3}$
The following neutralization reaction takes place:
$H_{2}SO_{4} + 2NaOH \rightarrow Na_{2}SO_{4} + 2H_{2}O$
The amount of sulfuric acid required is two times less than the amount of sodium - hydroxide, according to the stoichiometric coefficients. Therefore,
$n(H_{2}SO_{4}) = \frac{n(NaOH)}{2}=\frac{c(NaOH)\times V(NaOH)}{2}=\frac{0.454\frac{mol}{dm^3}\times 0.01158dm^3}{2}=0.00262866 mol$
Now we can easily obtain the volume of sulfuric acid solution:
$V(H_{2}SO_{4}) = \frac{n(H_{2}SO_{4})}{c(H_{2}SO_{4})}=\frac{0.00262866 mol}{2.28\frac{mol}{dm^3}}=0.00115dm^3 = 1.15ml$
