Question

$70ml,M/60so{l}^n$ of $KBr{O}_3$ in added to $Se{O}_3^{2-}$. Bromine evolved in removed by boiling & excess of $KBr{O}_3$ in back titrated with $12.5mlM/25so{l}^n$ of $NaAs{O}_2$. Calculate millilitre of $Se{O}_3^{2-}$.

Answer 1

$KBr{O}_3+Se{O}_3^{2-}⟶B{r}_2+Se{O}_4^{2-}$

$\stackrel{+5}{Br}⟶B{r}^o$ $nf=5$

$\stackrel{+4}{Se}{O}_3^{2-}⟶\stackrel{+6}{Se}{O}_4^{2-}$ $nf=2$

Thus, number of equivalent of $KBr{O}_3$ reacts with $Se{O}_3^{2-}$

$\underset{\left(KBr{O}_3\right)}{n_1\times n{f}_1}=\underset{\left(Se{O}_3^{2-}\right)}{n_2\times n{f}_2}\Rightarrow x\times 5=n_2\times 2⟶\left(1\right)$

Number of moles in $70ml$ of $\frac{M}{60}KBr{O}_3=\frac{70\times M}{1000\times 60}$ moles

Let $x$ moles of $KBr{O}_3$ reacts with $Se{O}_3^{2-}$

Thus, the excess i.e. $(\frac{70M}{60000}-x)$ will react with $NaAs{O}_2$

Let say $y$,

Now by using $n_3\times n{f}_1=n_4\times n{f}_4$

Excess $\left(KBr{O}_3\right)$

As $\stackrel{+3}{As}{O}_2^{-}⟶\stackrel{+5}{As}{O}_4^{3-}$

$n{f}_4=2$

Thus, $y\times 5=2\times n_4$

$\Rightarrow n_4=\frac{12.5L}{1000L}\times \frac{M}{25}$

$\Rightarrow n_4=\frac{0.5}{1000}M$

$\Rightarrow 5y=\frac{2\times 0.5}{1000}\times M$

$\Rightarrow y=\frac{1\times M}{5\times 1000}\Rightarrow$ excess of $KBr{O}_3$ moles

Thus, the moles of $KBr{O}_3$ reacts with $Se{O}_3^{2-}$

$\Rightarrow x=\frac{70M}{60\times 1000}-\frac{1}{5}\times \frac{M}{1000}=\frac{0.96M}{1000}$ moles

Now, put the value of $x$ in equation 1, we get

$\Rightarrow \frac{0.96M}{1000}\times 5=2\times n_2$

$\Rightarrow n_2=2.4\times {10}^{-3}M$ moles of $Se{O}_3^{2-}$

For $1M$ solution of $Se{O}_3^{2-}$

$V=2.4\times {10}^{-3}L=2.4ml$ of $1M$ solution of $Se{O}_3^{2-}$.
or $2.4$ millilitres of $Se{O}_3^{2-}$ are required.