Question

At certain temperature the ${H^{+}}$ ion concentration of ${H_{2}}O$ is $\begin{align*}4\times 10^{-7}M\end{align*}$, then the value of ${K_{w}}$ at the same temperature will be:
A.$\begin{align*} 10^{-14}M\end{align*}$
B.$\begin{align*} 4\times 10^{-14}M\end{align*}$
C.$\begin{align*} 1.6\times 10^{-13}M\end{align*}$
D.$\begin{align*} 4\times 10^{-7}M\end{align*}$

Answer 1

Hint: Dissociation constant or ionic product of water (${K_{w}}$) is defined as the product of the molar concentration of hydroxyl ion hydrogen ion concentration at a constant temperature.

Complete step-by-step answer:- The dissociation constant is defined as the product of the molar concentration of hydroxyl ion and hydrogen ion concentration at a constant temperature. It can be shown as;
$\begin{align*}K_{w} = [H^{+}][OH^{-}]\end{align*}$
- It is known that for pure water, hydrogen ion concentration should be equal to hydroxyl ion concentration.
$\begin{align*}[H^{+}]=[OH^{-}]\end{align*}$
Since water is neutral, for neutral solutions dissociation constant will be shown as
$\begin{align*}[H^{+}]=[OH^{-}]=4\times 10^{-7}M\end{align*}$
$\begin{align*}[H^{+}]^{2}=K_{w}\end{align*}$
Thus,
$\begin{align*}K_{w} =(4\times 10^{-7})^{2}=1.6\times 10^{-13}\end{align*}$

Additional information: The smaller the dissociation constant the weaker the acid or base would be. Dissociation constant depends on the way the substance would split apart. The concentration of the reactant will be equal to the concentration of each of the products formed.

Therefore, the correct answer is option (C).

Note: Dissociation constant is also known as ‘equilibrium constant’ or ‘ionization constant’ and is represented as ${K_{a}}$ for acids and as ${K_{b}}$ for bases. Larger the equilibrium constant the stronger the acid will be, and it will have more ${H^{+}}$ ion concentration at equilibrium. The equilibrium constant does not depend on the initial concentrations of reactants or products.