Question

The charge balance equation of species in $0.100$M acetic acid solution is given by
A. $\left[ {{{\text{H}}^{\text{ + }}}} \right]\, = \,\left[ {{\text{O}}{{\text{H}}^ - }} \right]$
B.$\left[ {{{\text{H}}^{\text{ + }}}} \right]\, = \,\left[ {{\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }} \right]$
C. $\left[ {{{\text{H}}^{\text{ + }}}} \right]\, = \,\left[ {{\text{O}}{{\text{H}}^ - }} \right] + \,\left[ {{\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }} \right]$
D. $2\left[ {{{\text{H}}^{\text{ + }}}} \right]\, = \,\left[ {{\text{O}}{{\text{H}}^ - }} \right] + \,\left[ {{\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }} \right]$

Answer 1

To solve this first calculate the number of moles of ${\text{KI}}$ in $200{\text{ ml}}$, $100{\text{ ml}}$ and $0.5{\text{ litre}}$ of $0.1{\text{ M}}$ solution of ${\text{KI}}$. Then calculate the number of moles of the given species that react with the calculated moles of ${\text{KI}}$. This can be solved by using the simple definition of molarity.

Complete step by step answer:
We are given that $0.1{\text{ M}}$ solution of ${\text{KI}}$ reacts with excess of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ and ${\text{KI}}{{\text{O}}_{\text{3}}}$ solutions, according to the equation as follows:
$5{{\text{I}}^ - } + {\text{IO}}_3^ - + 6{{\text{H}}^ + } \to {\text{3}}{{\text{I}}_2} + {\text{3}}{{\text{H}}_2}{\text{O}}$
The reaction can be written as follows:
$5{\text{KI}} + {\text{KI}}{{\text{O}}_3} + 3{{\text{H}}_2}{\text{S}}{{\text{O}}_4} \to {\text{3}}{{\text{I}}_2} + 3{{\text{K}}_2}{\text{S}}{{\text{O}}_4}{\text{ + 3}}{{\text{H}}_2}{\text{O}}$
The given equation is a balanced chemical equation.
$200{\text{ ml}}$ of ${\text{KI}}$ solution reacts with $0.004{\text{ mol}}$ ${\text{KI}}{{\text{O}}_{\text{3}}}$
We know that the molarity of a solution is the number of moles of solute per litre of solution. Thus,
${\text{Molarity}}\left( {\text{M}} \right) = \dfrac{{{\text{Number of moles}}\left( {{\text{mol}}} \right)}}{{{\text{Volume of solution}}\left( {{\text{litre}}} \right)}}$
Thus,
${\text{Number of moles}}\left( {{\text{mol}}} \right) = {\text{Molarity}}\left( {\text{M}} \right) \times {\text{Volume of solution}}\left( {{\text{litre}}} \right)$
Thus, the number of moles of ${\text{KI}}$ in $200{\text{ ml}} = 200 \times {10^{ - 3}}{\text{ litre}}$ of $0.1{\text{ M}}$ solution of ${\text{KI}}$ are:
${\text{Number of moles of KI}} = {\text{0}}{\text{.1 M}} \times 200 \times {10^{ - 3}}{\text{ litre}} = 0.02{\text{ mol}}$
Thus, $200{\text{ ml}}$ of $0.1{\text{ M}}$ solution of ${\text{KI}}$ corresponds to $0.02{\text{ mol}}$ of ${\text{KI}}$.
From the reaction, we can say that $5{\text{ mol KI}}$ reacts with $1{\text{ mol KI}}{{\text{O}}_3}$. Thus, the moles of ${\text{KI}}{{\text{O}}_3}$ that react with $0.02{\text{ mol}}$ of ${\text{KI}}$ are:
${\text{Moles of KI}}{{\text{O}}_3} = 0.02{\text{ mol KI}} \times \dfrac{{1{\text{ mol KI}}{{\text{O}}_3}}}{{5{\text{ mol KI}}}} = 0.004{\text{ mol KI}}{{\text{O}}_3}$
Thus, the statement ‘$200{\text{ ml}}$ of ${\text{KI}}$ solution reacts with $0.004{\text{ mol}}$ ${\text{KI}}{{\text{O}}_{\text{3}}}$’ is correct.
Thus, option (A) is correct.
$100{\text{ ml}}$ of ${\text{KI}}$ solution reacts with $0.006{\text{ mol}}$ of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$:
We know that the molarity of a solution is the number of moles of solute per litre of solution. Thus,
${\text{Molarity}}\left( {\text{M}} \right) = \dfrac{{{\text{Number of moles}}\left( {{\text{mol}}} \right)}}{{{\text{Volume of solution}}\left( {{\text{litre}}} \right)}}$
Thus,
${\text{Number of moles}}\left( {{\text{mol}}} \right) = {\text{Molarity}}\left( {\text{M}} \right) \times {\text{Volume of solution}}\left( {{\text{litre}}} \right)$
Thus, the number of moles of ${\text{KI}}$ in $100{\text{ ml}} = 100 \times {10^{ - 3}}{\text{ litre}}$ of $0.1{\text{ M}}$ solution of ${\text{KI}}$ are:
${\text{Number of moles of KI}} = {\text{0}}{\text{.1 M}} \times 100 \times {10^{ - 3}}{\text{ litre}} = 0.01{\text{ mol}}$
Thus, $100{\text{ ml}}$ of $0.1{\text{ M}}$ solution of ${\text{KI}}$ corresponds to $0.01{\text{ mol}}$ of ${\text{KI}}$.
From the reaction, we can say that $5{\text{ mol KI}}$ react with $3{\text{ mol }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}$. Thus, the moles of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ that react with $0.01{\text{ mol}}$ of ${\text{KI}}$ are:
${\text{Moles of }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4} = 0.01{\text{ mol KI}} \times \dfrac{{{\text{3 mol }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}}}{{5{\text{ mol KI}}}} = 0.006{\text{ mol }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}$
Thus, the statement ‘$100{\text{ ml}}$ of ${\text{KI}}$ solution reacts with $0.006{\text{ mol}}$ of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$’ is correct.
Thus, option (B) is correct.
$0.5{\text{ litre}}$ of ${\text{KI}}$ solution reacts with $0.005{\text{ mol}}$ of ${{\text{I}}_{\text{2}}}$:
We know that the molarity of a solution is the number of moles of solute per litre of solution. Thus,
${\text{Molarity}}\left( {\text{M}} \right) = \dfrac{{{\text{Number of moles}}\left( {{\text{mol}}} \right)}}{{{\text{Volume of solution}}\left( {{\text{litre}}} \right)}}$
Thus,
${\text{Number of moles}}\left( {{\text{mol}}} \right) = {\text{Molarity}}\left( {\text{M}} \right) \times {\text{Volume of solution}}\left( {{\text{litre}}} \right)$
Thus, the number of moles of ${\text{KI}}$ in $0.5{\text{ litre}}$ of $0.1{\text{ M}}$ solution of ${\text{KI}}$ are:
${\text{Number of moles of KI}} = {\text{0}}{\text{.1 M}} \times 0.5{\text{ litre}} = 0.05{\text{ mol}}$
Thus, $0.5{\text{ litre}}$ of $0.1{\text{ M}}$ solution of ${\text{KI}}$ corresponds to $0.05{\text{ mol}}$ of ${\text{KI}}$.
From the reaction, we can say that $5{\text{ mol KI}}$ produce $3{\text{ mol }}{{\text{I}}_2}$. Thus, the moles of ${{\text{I}}_2}$ that are produced by with $0.05{\text{ mol}}$ of ${\text{KI}}$ are:
${\text{Moles of }}{{\text{I}}_2} = 0.05{\text{ mol KI}} \times \dfrac{{{\text{3 mol }}{{\text{I}}_2}}}{{5{\text{ mol KI}}}} = 0.03{\text{ mol }}{{\text{I}}_2}$
Thus, the statement ‘$0.5{\text{ litre}}$ of ${\text{KI}}$ solution reacts with $0.005{\text{ mol}}$ of ${{\text{I}}_{\text{2}}}$’ is incorrect.
Thus, option (C) is not correct.
Equivalent weight of ${\text{KI}}{{\text{O}}_{\text{3}}}$ is equal to $\left( {\dfrac{{{\text{Molecular weight}}}}{5}} \right)$:
The equation that gives the relationship between the equivalent weight and the molecular weight is,
${\text{Equivalent weight}} = \dfrac{{{\text{Molecular weight}}}}{{{\text{Valency factor}}}}$
The valency factor is the change in oxidation state.
We are given the reaction,
$5{\text{KI}} + {\text{KI}}{{\text{O}}_3} + 3{{\text{H}}_2}{\text{S}}{{\text{O}}_4} \to {\text{3}}{{\text{I}}_2} + 3{{\text{K}}_2}{\text{S}}{{\text{O}}_4}{\text{ + 3}}{{\text{H}}_2}{\text{O}}$
In the reaction, ${\text{KI}}{{\text{O}}_3}$ is reduced to ${{\text{I}}_2}$. The oxidation state of iodine i.e. ${\text{I}}$ in ${\text{KI}}{{\text{O}}_3}$ is $+ 5$ and in ${{\text{I}}_2}$ is 0. Thus, the change in oxidation state is 5. Thus, the valency factor for ${\text{KI}}{{\text{O}}_3}$ is 5.
Thus,
${\text{Equivalent weight of KI}}{{\text{O}}_3} = \dfrac{{{\text{Molecular weight of KI}}{{\text{O}}_3}}}{{\text{5}}}$
Thus, the statement ‘equivalent weight of ${\text{KI}}{{\text{O}}_{\text{3}}}$ is equal to $\left( {\dfrac{{{\text{Molecular weight}}}}{5}} \right)$’ is correct.

Thus, the correct option is A,B,D.

Note: Remember that the equation for the reaction must be balanced. Unbalanced chemical equations can lead to incorrect answers. Also, remember to calculate the correct valency factor. The equivalent weight is the ratio of molecular weight to the valency factor. For acids, valency factor is the number of replaceable hydrogen ions and for bases, valency factor is the number of replaceable hydroxide ions.