Question

(i) What is the mass of sodium bromate and molarity of the solution necessary to prepare $85.4mL$ of $0.672N$ solution when the half reaction is ${Br{O}_3}^{-}+6H^{+}+6e^{-}\to B{r}^{-}+3H_{2}O$?
(ii) What would be the mass as well as molarity if the half cell reaction is $2{Br{O}_3}^{-}+12H^{+}+10e^{-}\to B{r}_2+6H_{2}O$?

Answer 1

(i) Molecular mass of $NaBr{O}_3=23+80+(3\times 16)=151$
Each bromate ion takes-up $6$ electrons; therefore,
Eq. mass of $NaBr{O}_3=\frac{Mol.mass}{6}=\frac{151}{6}$
Amount of $NaBr{O}_3$ in $85.5mL0.672N$ solution
$=\frac{0.672}{1000}\times \frac{15!}{6}\times 85.5=1.446g$
$Molarity=\frac{Normality}{n}=\frac{0.672}{6}=0.112M$
(ii) Each bromate ion takes-up $5$ electrons; therefore,
Eq. mass of $NaBr{O}_3=\frac{Mol.mass}{5}=\frac{151}{5}$
Amount of $NaBr{O}_3$ in $85.5mL0.672N$ solution
$=\frac{151}{2}\times \frac{0.672}{1000}\times 85.5$
$=1.7352g$
$Molarity=\frac{Normality}{n}=\frac{0.672}{4}-0.1344M$.