Question

If pure water has ${\text{pKw = 13}}{\text{.36}}$ at ${\text{50}}^\circ {\text{C}}$ the pH of pure water will be:
(A) $6.68$
(B) $7.0$
(C) $7.13$
(D) $6.0$

Answer 1

The equilibrium constant, dissociation constant or the ionization constant for the water at the equilibrium is represented by ${\text{Kw}}$ . This ionization constant changes with the change in the temperature of the system. It is only temperature dependent.

Complete step by step answer:
In chemical equilibrium the Kw represents the equilibrium constant or the dissociation constant or the ionization constant for water. Normally the value of dissociation constant is taken as ${10^{ - 7}}$ .
The other term pKw which is basically the sum of acidic and basic values at a particular temperature (pKw= pH + pOH). For pure water at room temperature the pKw value is considered 14 in the chemical equilibrium.
So as we know that,
pKw= pH + pOH
And for a neutral solution the values of concentrations of ${{\text{H}}^{\text{ + }}}{\text{ and O}}{{\text{H}}^{\text{ - }}}$ ions are always equal.
Therefore, pH=pOH ; for neutral solution
By using this condition in the above formula we get;
$pKw = {\text{ 2}} \times pH$
And we have given that the pure water has ${\text{pKw = 13}}{\text{.36}}$ at ${\text{50}}^\circ {\text{C}}$
so by putting the value in the above equation we get;
$pH = \dfrac{{13.36}}{2} = 6.68$
Therefore if pure water has ${\text{pKw = 13}}{\text{.36}}$ at ${\text{50}}^\circ {\text{C}}$ then the pH of pure water will be 6.68.
So option (1) is the correct answer.

Note:
Kw is known auto pyrolysis constant of water because it automatically dissociates itself. The Kw or auto pyrolysis constant for the water at $25^\circ {\text{C}}$ is always equal to ${10^{ - 14}}$ . The relation between the Kw and pKw is; ( $pKw = - \log Kw$ ).