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Question: What are the molar concentrations of \[\left[ {{H^ + }} \right]\] and \(\left[ {O{H^ - }} \right]\) ...

What are the molar concentrations of [H+]\left[ {{H^ + }} \right] and [OH−]\left[ {O{H^ - }} \right] in pure water at 25∘C25^\circ C ?

Explanation

Solution

Water undergoes auto-protolysis of self-ionization to a small extent. This leads to the fact on writing down the reaction of self-ionization of water, we can calculate the equilibrium constant Kw{K_w} which is equal to the product of [H+]\left[ {{H^ + }} \right] and [OH−]\left[ {O{H^ - }} \right]. Since it has been established that the value of Kw = 1×10−14  mol2L−2 at  25∘C,{K_w}\, = \,1 \times {10^{ - 14\,}}\,mo{l^2}{L^{ - 2}}\,{\text{at }}\,25^\circ C, it is possible to calculate the values of [H+]\left[ {{H^ + }} \right] and [OH−]\left[ {O{H^ - }} \right],
[H3O+] = [OH−] = 1×10−7mol L−1\left[ {{H_3}{O^ + }} \right]\, = \,\left[ {O{H^ - }} \right]\, = \,1 \times {10^{ - 7}}mol\,{L^{ - 1}}

Complete answer:
Water undergoes self-ionization to a small extent as:
The equilibrium constant for this ionization is given by:
K = [H3O+][OH−][H2O]2K\, = \,\dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right]}}{{{{\left[ {{H_2}O} \right]}^2}}}
Or K[H2O]2 = [H3O+][OH−]K{\left[ {{H_2}O} \right]^2}\, = \,\left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right]
Since the degree of ionization of water is very small, [H2O]\left[ {{H_2}O} \right] may be regarded as practically constant. So, we use another constant instead, called Kw{K_w} in the place of K[H2O]2K{\left[ {{H_2}O} \right]^2}:
Kw = [H3O+][OH−]{K_w}\, = \,\left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right]
Where Kw{K_w} is called the “Ionic product of water”.
“The ionic product of water (Kw)\left( {{K_w}} \right) at a given temperature is defined as the product of the molar concentrations of [H3O+]\left[ {{H_3}{O^ + }} \right] and [OH−]\left[ {O{H^ - }} \right] ions at that temperature.”
As we know, Kw{K_w} has a value of 1×10−14  mol2L−2 (or 1×10−14  M2)  at  25∘C1 \times {10^{ - 14\,}}\,mo{l^2}{L^{ - 2}}\,({\text{or}}\,1 \times {10^{ - 14\,}}\,{M^2})\,\,{\text{at }}\,25^\circ C .
However, the value of Kw{K_w} increases with increase in temperature. It shows increased values at 30∘C30^\circ C and at 40∘C40^\circ C :
At 30∘C,30^\circ C, Kw = 1.5× 10−14 mol2L−2{K_w}\, = \,1.5 \times \,{10^{ - 14}}\,mo{l^2}{L^{ - 2}}
At 40∘C,40^\circ C, Kw = 3× 10−14 mol2L−2{K_w}\, = \,3 \times \,{10^{ - 14}}\,mo{l^2}{L^{ - 2}}
Since we are only concerned with the value of Kw{K_w} at 25∘C25^\circ C ,
Kw = [H3O+][OH−] = 1×10−14mol2L−2{K_w}\, = \,\left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right]\, = \,1 \times {10^{ - 14}}mo{l^2}{L^{ - 2}}
Thus, in pure water or neutral solution,
[H3O+] = [OH−] = 1×10−7mol L−1\left[ {{H_3}{O^ + }} \right]\, = \,\left[ {O{H^ - }} \right]\, = \,1 \times {10^{ - 7}}mol\,{L^{ - 1}}
Hence, molar concentrations of [H+]\left[ {{H^ + }} \right] and [OH−]\left[ {O{H^ - }} \right] in pure water at 25∘C25^\circ C is the same value, which is 1×10−7mol L−11 \times {10^{ - 7}}mol\,{L^{ - 1}}.

Note:
[H+]\left[ {{H^ + }} \right] can be denoted as [H3O+] \left[ {{H_3}{O^ + }} \right]\, since they are both representations for the proton. Kw{K_w} is a constant at constant temperatures, hence the product [H3O+][OH−]\,\left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right] remains constant too. Another case to be noted is that whether the solution is neutral, acidic or basic, the value of Kw{K_w} is a constant at constant temperatures.