Question

Write balanced net ionic equation for the following reactions in acidic solution.

$ClO{{}_3}^{-}\left(aq\right)+A{s}_2{S}_3\left(s\right)\to C{l}^{-}\left(aq\right)+H_{2}AsO{{}_4}^{-}\left(aq\right)+HSO{{}_4}^{-}\left(aq\right)$

• A. $14ClO{{}_3}^{-}\left(aq\right)+A{s}_2{S}_3\left(s\right)+18H_{2}O\left(l\right)\to 14C{l}^{-}\left(aq\right)+6H_{2}AsO{{}_4}^{-}\left(aq\right)$ $+9HSO{{}_4}^{-}\left(aq\right)+15H^{+}\left(aq\right)$
• B. $14ClO{{}_3}^{-}\left(aq\right)+2A{s}_2{S}_3\left(s\right)+20H_{2}O\left(l\right)\to 14C{l}^{-}\left(aq\right)+8H_{2}AsO{{}_4}^{-}\left(aq\right)$$+9HSO{{}_4}^{-}\left(aq\right)+16H^{+}\left(aq\right)$
• C. $7ClO{{}_3}^{-}\left(aq\right)+A{s}_2{S}_3\left(s\right)+9H_{2}O\left(l\right)\to 7C{l}^{-}\left(aq\right)+4H_{2}AsO{{}_4}^{-}\left(aq\right)$$+9HSO{{}_4}^{-}\left(aq\right)+15H^{+}\left(aq\right)$
• D. None of the above

Answer 1

Answer: (A)

$14ClO{{}_3}^{-}\left(aq\right)+A{s}_2{S}_3\left(s\right)+18H_{2}O\left(l\right)\to 14C{l}^{-}\left(aq\right)+6H_{2}AsO{{}_4}^{-}\left(aq\right)$ $+9HSO{{}_4}^{-}\left(aq\right)+15H^{+}\left(aq\right)$

The unbalanced chemical equation is ,

$ClO{{}_3}^{-}\left(aq\right)+A{s}_2{S}_3\left(s\right)\to C{l}^{-}\left(aq\right)+H_{2}AsO{{}_4}^{-}\left(aq\right)+HSO{{}_4}^{-}\left(aq\right)$.

Balance $As$ and $S$ atoms.

$ClO{{}_3}^{-}\left(aq\right)+A{s}_2{S}_3\left(s\right)\to C{l}^{-}\left(aq\right)+2H_{2}AsO{{}_4}^{-}\left(aq\right)+3HSO{{}_4}^{-}\left(aq\right)$

The oxidation number of $Cl$ decreases from $+5$ to $-1$. Total decrease in the oxidation number is $6$.

The oxidation number of $As$ increases from $+3$ to $+5$. Total increase in the oxidation number for $2As$ atoms is $4$.

The oxidation number of $S$ increases from $-2$ to $+6$. Total increase in the oxidation number for $3S$ atoms is $24$.

Total increase in the oxidation number is $28$.

To balance total increase in the oxidation number with total decrease in the oxidation number, multiply chlorine containing species with $14$ and $As$ and $S$ containing species with $3$.

$14ClO{{}_3}^{-}\left(aq\right)+3A{s}_2{S}_3\left(s\right)\to 14C{l}^{-}\left(aq\right)+6H_{2}AsO{{}_4}^{-}\left(aq\right)+9HSO{{}_4}^{-}\left(aq\right)$

To balance O atoms, add 18 water molecules on LHS.

$14ClO{{}_3}^{-}\left(aq\right)+3A{s}_2{S}_3\left(s\right)+18H_{2}O\left(l\right)\to 14C{l}^{-}\left(aq\right)+6H_{2}AsO{{}_4}^{-}\left(aq\right)+9HSO{{}_4}^{-}\left(aq\right)$

To balance hydrogen atoms, add 15 protons on the RHS.

$14ClO{{}_3}^{-}\left(aq\right)+3A{s}_2{S}_3\left(s\right)+18H_{2}O\left(l\right)\to 14C{l}^{-}\left(aq\right)+6H_{2}AsO{{}_4}^{-}\left(aq\right)+9HSO{{}_4}^{-}\left(aq\right)+15H^{+}\left(aq\right)$

This is the balanced chemical equation.