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

Hint: 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 $ ).