Answer
$V(CH_{3}COOH)=31.25ml$
Work Step by Step
$c(CH_{3}COOH)=0.145\frac{mol}{dm^{3}}$
$V(Ba(OH)_{2})=21.58ml=0.02158dm^{3}$
$c(Ba(OH)_{2})=0.105\frac{mol}{dm^{3}}$---$V(CH_{3}COOH)=?$
The neutralization reaction of $Ba(OH)_{2}$ by $CH_{3}COOH$:
$2CH_{3}COOH+Ba(OH)_{2} \rightarrow (CH_{3}COO)_{2}Ba+2H_{2}O$
We can observe that the amount of acetic acid required for the complete neutralization is twice the amount of barium hydroxide (according to the reaction's coefficients).
$n(CH_{3}COOH)=2n(Ba(OH)_{2})=2c(Ba(OH)_{2})\times V(Ba(OH)_{2})=2\times 0.105\frac{mol}{dm^{3}}\times 0.02158dm^{3}=4.5318\times 10^{-3}mol$
Now, when we have calculated the amount of acetic acid required for the complete neutralization, we can obtain the required volume of the solution by using its concentration:
$V(CH_{3}COOH)=\frac{n(CH_{3}COOH)}{c(CH_{3}COOH)}=\frac{4.5318\times 10^{-3}mol}{0.145\frac{mol}{dm^{3}}}=0.03125 dm^{3}=31.25ml$
