Answer
$c(KI)=5.4895\frac{mol}{dm^{3}}$
Work Step by Step
$V(HI)=55.5ml=0.0555dm^{3}$
$c(HI)=8.99\frac{mol}{dm^{3}}$
$V(KOH)=35.4ml=0.0354dm^{3}$
$c(HI)=14.1\frac{mol}{dm^{3}}$
$n(HI)=c(HI)\times V(HI)=8.99\frac{mol}{dm^{3}}\times 0.0555dm^{3}=0.498945mol$
$n(KOH)=c(KOH)\times V(KOH)=14.1\frac{mol}{dm^{3}}\times 0.0354dm^{3}=0.49914mol$
We have shown that $n(HI)\approx n(KOH)$.
The following reaction occurs:
$HI+KOH \rightarrow KI + H_{2}O$
$n(KI)=n(HI)=n(KOH)=0.499 mol$
$V_{tot}=V(HI)+V(KOH)=90.9ml=0.0909dm^{3}$
$c(KI)=\frac{n(KI)}{V_{tot}}=\frac{0.499mol}{0.0909dm^{3}}=5.4895\frac{mol}{dm^{3}}$
