Question

What is the effect of temperature on the ionic product of water? How will it change the pH value of a neutral solution?

Answer 1

Ionic product is the dissociation of a water molecule into hydrogen ions and hydroxyl ions. When the temperature increases the number of hydrogen ions and hydroxyl ions also increases. We know that the pH is equal to the negative logarithm of the concentration of hydrogen ions.

Complete step by step answer:
Pure water is a poor conductor of electricity. This shows that water is a weak electrolyte, i.e., it is ionized to a very small extend as:
${{H}_{2}}O\rightleftharpoons {{H}^{+}}+O{{H}^{-}}$
When we apply the law of chemical equilibrium to the above equilibrium reaction, we get:
$K=\dfrac{[{{H}^{+}}][O{{H}^{-}}]}{[{{H}_{2}}O]}$
Where K is the dissociation constant of water.
So, this equation can be written as:
$K[{{H}_{2}}O]=[{{H}^{+}}][O{{H}^{-}}]$
So, $K[{{H}_{2}}O]={{K}_{w}}$
${{K}_{w}}=[{{H}^{+}}][O{{H}^{-}}]$
So, ${{K}_{w}}$is another constant, called the ionic product of water.
So, we can say that the ionic product of water is the product of the molar concentration of hydrogen ions $[{{H}^{+}}]$and hydroxyl ions$[O{{H}^{-}}]$.
It must be noted that ionic water is constant only at a constant temperature. Its value at ${{25}^{\circ }}C$or 298 K is found to be$1.008\text{ x 1}{{\text{0}}^{-14}}$.
${{K}_{w}}=1.008\text{ x 1}{{\text{0}}^{-14}}$
Effect of temperature of ${{K}_{w}}$The ionic product of water (${{K}_{w}}$) increases with increases in temperature. This is obviously because of the fact that with an increase of temperature, the degree of ionization of water increases. In other words, the ${{H}_{2}}O$dissociate into ${{H}^{+}}$and$O{{H}^{-}}$. Thus the concentration of hydrogen and hydroxyl ions increases.
We know that the pH is equal to the negative logarithm of the concentration of hydrogen ions.
$pH=-\log [{{H}^{+}}]$
So, when the temperature increases the concentration of hydrogen ions increases. According to the formula,$pH=-\log [{{H}^{+}}]$, the increase in hydrogen ions the pH of the solution will decrease.

Note: It must be noted that the dissociation/ionization constant of water (K) is different from the ionic product of water. These are related as: $K=\dfrac{{{K}_{w}}}{55.55}$(because the concentration of water is 55.55) and the ionic product of water is ${{10}^{-14}}$, so
$K=\dfrac{{{10}^{-14}}}{55.55}=1.8\text{ x 1}{{\text{0}}^{-16}}$
This value is less than the ionic product of water which shows that only a few molecules of ${{H}_{2}}O$ undergo dissociation.
The relation: ${{\text{K}}_{a}}\text{ x }{{\text{K}}_{b}}\text{ = }{{\text{K}}_{w}}$, is used to solve numerical. Where ${{K}_{a}}$is the ionization constant of acid and ${{K}_{b}}$is the ionization constant of conjugate base.