Question

In the following reaction:
$O{H}^{-}+S_2{O}_3^{2-}+B{r}_2\to S{O}_4^{2-}+B{r}^{-}$, starting with $0.15$ mol of $B{r}_2,0.01$ and of $S_2{O}_3^{2-}$ and $0.4$ mol of $O{H}^{-}$ ions, if $x$ moles of $O{H}^{-}$ ions are left in the solution after the reaction is complete, then calculate the value of $10x$.

• A. $3$
• B. $4$
• C. $2$
• D. None of these

Answer 1

Answer: (A)

$3$

Balance the equation in basic medium
$10O{H}^{-}+S_2{O}_3^{2-}\to 2S{O}_4^{2-}+5H_{2}O+8e^{-}$

[ $2e^{-}+B{r}_2\to 2B{r}^{⊝}]\times 4$
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$4B{r}_2+S_2{O}_3^{2-}+10\stackrel{⊝}{O}H\to 2S{O}_4^{2-}+8B{r}^{⊝}+5H_{2}O$
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$\left[\begin{array}{c}Moles\\ before\\ reaction\end{array}\right]$ $\left(0.15\right)$ $\left(0.01\right)$ $\left(0.4\right)$ $\left(0\right)$ $\left(0\right)$ $\left(0\right)$

$\left[\begin{array}{c}Moles\\ reacted\\ and\\ formed\end{array}\right]$ $(0.01\times 4)$ $\left(0.01\right)$ $(0.01\times 10)$ $(0.01\times 2)$ $(0.01\times 8)$ $(0.01\times 5)$

$\left[\begin{array}{c}Moles\\ left\end{array}\right]$ $\underset{(=0.11)}{(0.15-0.04)}$ $\underset{(=0.3)}{(0.4-0.1)}$ $\left(0.02\right)$ $\left(0.08\right)$ $\left(0.05\right)$

So, molar ratio of combination is:

$B{r}_2:S_2{O}_3^{2-}:\stackrel{⊝}{O}H::4:1:10$

$\left[\stackrel{⊝}{O}H\right]$ left after reaction $=0.3$ mol

So, $10\times x=10\times 0.3=3$