Answer
$c(BaI_{2})=0.0376\frac{mol}{dm^{3}}$
Work Step by Step
$V(Ba(OH)_{2})=7.5ml=0.0075dm^{3}$
$c(Ba(OH)_{2})=0.135\frac{mol}{dm^{3}}$
$V(HI)=19.4ml=0.0194dm^{3}$
$c(HI)=0.104\frac{mol}{dm^{3}}$
$n(Ba(OH)_{2})=c(Ba(OH)_{2})×V(Ba(OH)_{2})=0.135\frac{mol}{dm^{3}}×0.0075dm^{3}=1.0125\times 10^{-3}mol$
$n(HI)=c(HI)×V(HI)=0.104\frac{mol}{dm^{3}}×0.0194dm^{3}=2.0176\times 10^{-3}mol$
We have shown that $n(HI)≈2n(Ba(OH)_{2})$.
The following reaction occurs:
$Ba(OH)_{2}+2HI→BaI_{2}+2H_{2}O$
$Ba(OH)_{2}$ and $HI$ react in $1:2$ ratio.
$n(BaI_{2})=n(Ba(OH)_{2})=1.0125\times 10^{-3}mol$
$V_{tot}=V(Ba(OH)_{2})+V(HI)=26.9ml=0.0269dm^{3}$
$\implies c(BaI_{2})=\frac{n(BaI_{2})}{V_{tot}}=0.0376\frac{mol}{dm^{3}}$
