Question

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

Answer 1

Hint: 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 ${K_w}$ which is equal to the product of \[\left[ {{H^ + }} \right]\] and $\left[ {O{H^ - }} \right]$. Since it has been established that the value of ${K_w}\, = \,1 \times {10^{ - 14\,}}\,mo{l^2}{L^{ - 2}}\,{\text{at }}\,25^\circ C,$ it is possible to calculate the values of \[\left[ {{H^ + }} \right]\] and $\left[ {O{H^ - }} \right]$,
$\left[ {{H_3}{O^ + }} \right]\, = \,\left[ {O{H^ - }} \right]\, = \,1 \times {10^{ - 7}}mol\,{L^{ - 1}}$

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

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