Question

Assertion:
(A)$\to$When water is heated to ${80}^{\circ }C$ then concentrations $\left[H^{+}\right]and\left[O{H}^{-}\right]$ will increase.
Reason:
(R)$\to K_w$ increases with decrease in temperature.

• A. Both, A and R, are true and R is the correct explanation of A
• B. Both, A and R, are true but R is not the correct explanation of A
• C. A is true but R is false
• D. Both A and R are false

Answer 1

The correct option is C A is true but R is false

$K_w$​ increases with increase in temperature as with the increase in temperature more no. of molecules dissociate and ionic product increases.
$H_{2}O\left(l\right)\rightleftharpoons H^{+}(aq.)+O{H}^{-}(aq.)K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$
At equilibrium, $\left[H^{+}\right]=\left[O{H}^{-}\right]$
If $K_w$ increases, then , concentration oh $H^{+}andO{H}^{-}$ also increases.



Theory :

Ionic Product of water:
$H_{2}O\left(l\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}(aq.)+O{H}^{-}(aq.)$ or $2H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$
as per laws of mass action : $K_c=\frac{\left[H^{+}\right]\left[O{H}^{-}\right]}{\left[H_{2}O\right]}$
In pure water,and in dilute aqueous solutions The molar concentration of water is essentially a constant, with molarity of $55.55mol/L$.
Density of $H_{2}O=1g/ml$
or Mass of $H_{2}O$ in g = Volume of $H_{2}O$ in ml = 1g or 1 ml
Mass of $H_{2}O$ in 1L of water= 1000 g

$Molarity\left(M\right)=\frac{molesofwater}{Volumeofsolution}$ $M=\frac{\frac{1000}{18}}{1}=55.55mol/L$
The ratio of dissociated water to that of undissociated water can be given as :
$\frac{{10}^{-7}}{55}=1.8\times {10}^{-9}$.Thus, equilibrium lies mainly towards undissociated water.
Therefore, $K_c\times \left[H_{2}O\right]=\left[H^{+}\right]\left[O{H}^{-}\right]$ $K_c\times H_{2}O$ is a constant which is also called ionic product of water $K_w$.

Effect of temperature on $K_w$:

The ionic product of water $K_w$ increases with increase of temperature. The dissociation of water molecule into it's ions is an endothermic reaction.
$2H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+O{H}^{-}(aq.)\Delta H>0$

By Le Chatelier's principle, increase in the temperature will shift the equilibrium in forward direction. So more ions are formed, and $K_w$ increases.
$K_w\propto T$
Note:- Ionic product of water is always a constant whatever may be dissolved in water. As it is an equilibrium constant, it will depend only on temperature.