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

Ionic Product Of Water

Category : JEE Main & Advanced

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

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