JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Ionic Product Of Water

Water is a weak electrolyte and undergoes selfionistion to a small extent.

“The product of concentrations of \[{{H}^{+}}\] and \[O{{H}^{-}}\] ions in water at a particular temperature is known as ionic product of water.” It is designated as \[{{K}_{w}}\].

\[{{H}_{2}}O\] ? \[{{H}^{+}}+O{{H}^{-}}\]; \[\Delta H=+57.3\ kJ{{M}^{-1}}\]

\[K=\frac{[{{H}^{+}}][O{{H}^{-}}]}{[{{H}_{2}}O]}\];\[K[{{H}_{2}}O]=[{{H}^{+}}][O{{H}^{-}}]\];\[{{K}_{w}}=[{{H}^{+}}][O{{H}^{-}}]\]

The value of \[{{K}_{w}}\] increases with the increase of temperature, i.e., the concentration H⁺ and OH^– ions increases with increase in temperature.

The value of \[{{K}_{w}}\] at \[{{25}^{o}}C\] is \[1\times {{10}^{-14}}\]mole/litre. Since pure water is neutral in nature, \[{{H}^{+}}\] ion concentration must be equal to \[O{{H}^{-}}\] ion concentration.

\[[{{H}^{+}}]=[O{{H}^{-}}]=x\] or \[[{{H}^{+}}][O{{H}^{-}}]={{x}^{2}}=1\times {{10}^{-14}}\] or \[x=1\times {{10}^{-7}}M\] or  \[[{{H}^{+}}]=[O{{H}^{-}}]=1\times {{10}^{-7}}\ mole\ litr{{e}^{-1}}\]

This shows that at \[{{25}^{o}}C\], in 1 litre only \[{{10}^{-7}}\] mole of water is in ionic form out of a total of approximately 55.5 moles.

 Thus when, \[[{{H}^{+}}]=[O{{H}^{-}}]\]; the solution is neutral

                     \[[{{H}^{+}}]>[O{{H}^{-}}]\]; the solution is acidic

                       \[[{{H}^{+}}]<[O{{H}^{-}}]\]; the solution is basic