Question

From the value of ionic product of water at 25°C find out the concentration of hydroxyl ions?

Answer 1

Hydroxyl ion concentration= $[OH^{-}]= 10^{-7}$

Explanation:

At a particular temperature, product of concentration of $H^{+}$ and $OH^{-}$ ion in water is known as ionic product of water.

Dissociation of water is occurs as following

                             $H_{2}O\rightleftharpoons H^{+}+OH^{-}$

According to the law of chemical equilibrium-

                                  $K=\frac{[H^{+}][OH^{-}]}{[H_{2}O]}$

                             $K[H_{2}O]=[H^{+}][OH^{-}]$

Water is a very weak electrolyte so dissociate at very small extent therefore $[H_{2}O]$ will be constant. This gives the another constant $K_{w}$.

                                   $K_{w} =[H^{+}][OH^{-}]$     ......(i)

$K_{w}$ is the ionic product of water.

The value of $K_{w}$ vary with temperature. On increase in temperature, the value of $K_{w}$ increases due to increase in ionisation.

At  $25^{0}C$              $K_{W}=1\times 10^{-14}$          .....(ii)

From equation (i) and (ii)

                        $[H^{+}][OH^{-}]=1\times 10^{-14}$

Nature of pure water is neutral, so the concentrations of  $[H^{+}] and [OH^{-}]$ is equal.

So,                      $[H^{+}]= [OH^{-}]$    

Suppose,            $[H^{+}]= [OH^{-}]=x$

or                       $[H^{+}] [OH^{-}]=x^{2}$$=1\times 10^{-14}$

or                        $x=1\times 10^{-7}$

or                       $[H^{+}]= [OH^{-}]=1\times 10^{-7}$

So the concentration of hydroxyl ion $[OH^{-}]$ at $25^{0}C$ is $10^{-7}$.