Answer
$c(H_{3}PO_{4}) = 0.0539M$
$V(H_{3}PO_{4}) = 32.5ml$
Work Step by Step
$m(H_{3}PO_{4}) = 0.978g$
$V = 185ml = 0.185 dm^3$
----$c(H_{3}PO_{4}) = \frac{n(H_{3}PO_{4})}{V}=\frac{m(H_{3}PO_{4})}{M(H_{3}PO_{4})V}=\frac{0.978g}{98\frac{g}{mol}\times 0.185dm^3}=0.0539M$
The following is the reaction of neutralization of $H_{3}PO_{4}$ by $NaOH$:
$H_{3}PO_{4} + 3NaOH \rightarrow Na_{3}PO_{4} + 3H_{2}O$
Considering stoichiometric coefficients, we can observe that the amount of $H_{3}PO_{4}$ is three times less than the amount of sodium - hydroxide required for the reaction.
$n(H_{3}PO_{4}) = \frac{n(NaOH)}{3} = \frac{c(NaOH)\times V(NaOH)}{3} = \frac{0.454 \frac{mol}{dm^3}\times 0.01158dm^3}{3}=0.00175244mol$
Knowing the concentration and amount of $H_{3}PO_{4}$, we can easily obtained the required volume:
$V(H_{3}PO_{4}) = \frac{n(H_{3}PO_{4})}{c(H_{3}PO_{4})} = \frac{0.00175244mol}{0.0539\frac{mol}{dm^3}}=0.0325dm^3 = 32.5ml$
