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# What is the product ${H^ + }$ and $O{H^ - }$ ions concentration at room temperature?(A) ${10^{ - 14}}$ (B) ${10^{ - 12}}$(C) $2 \times {10^{ - 6}}$ (D) $2 \times {10^{ - 7}}$  Verified
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Hint: By using the formula of ionic product of water, we can easily find the product of ${H^ + }$ and $O{H^ - }$ ions concentration at room temperature.
${K_W} = [{H^ + }][O{H^ - }]$

Complete Solution :
Due to the electrical conductivity of water, it undergoes dissociation. Water being a weak electrolyte undergoes only partial dissociation. Dissociation of pure water to a small extent into ${H^ + }$ and $O{H^ - }$ ions by itself is called self-dissociation.
Equilibrium reaction of the pure water is given below:
${H_2}O \rightleftharpoons {H^ + } + O{H^ - }$
Dissociation constant of the acid is
$K = \dfrac{{[{H^ + }][O{H^ - }]}}{{{H_2}O}}$……. (1)
Concentration of the unionized ions is the same as the starting concentration.
${\text{[}}{{\text{H}}_{\text{2}}}{\text{O] = constant}}$
Therefore, equation (1) becomes
${K_W} = [{H^ + }][O{H^ - }]$…… (2)
Equation (2) gives ionic products of water.

Ionic product of pure water at ${25^o}C$ is
${K_W} = [{H^ + }][O{H^ - }] = 1 \times {10^{ - 14}}$……... (3)
In pure water, hydrogen ion concentration $[{H^ + }]$ is equal to the hydroxide ion concentration $[O{H^ - }]$.
i.e. $[{H^ + }] = [O{H^ - }]$

The concentration of each ion can be calculated from the equation for the ionic product of water.
We can take the concentration of $[{H^ + }]$ and $[O{H^ - }]$as $x$.
Substituting $x$ into equation (3) gives
${K_W} = [{H^ + }][O{H^ - }] = 1 \times {10^{ - 14}}$
${K_W} = {x^2} = 1 \times {10^{ - 14}}$
$x = 1 \times {10^{ - 7}}$
Therefore, $[{H^ + }] = [O{H^ - }] = 1 \times {10^{ - 7}}$
Hence, product of ${H^ + }$ and $O{H^ - }$ ions concentration at room temperature is
$[{H^ + }][O{H^ - }] = 1 \times {10^{ - 7}} \times 1 \times {10^{ - 7}} = 1 \times {10^{ - 14}} = {10^{ - 14}}$
So, the correct answer is “Option A”.