Question

The ionization constant of water is \[1.1 \times {10^{ - 16}}\]. What is the ionic product of water?
A.\[1.1 \times {10^{ - 16}} \times 6.023 \times {10^{23}}\]
B.\[\dfrac{{1.1 \times {{10}^{ - 16}}}}{{0.023 \times {{10}^{23}}}}\]
C.\[1.1 \times {10^{ - 16}} \times 55.4\]
D.\[\dfrac{{1.1 \times {{10}^{ - 16}}}}{{{{10}^{ - 14}}}}\]

Answer 1

Hint: We know that pure water is a bad conductor of electricity. Water is a weak electrolyte meaning that it is ionized to a very small extent.
\[{H_2}O \rightleftarrows {H^ + } + O{H^ - }\]
\[{H_2}O + {H_2}O \rightleftarrows {H_3}{O^ + } + O{H^ - }\]
This ionization is known as self-ionization of water.

Complete answer:
We have,
\[{H_2}O \rightleftarrows {H^ + } + O{H^ - }\]
\[{H_2}O + {H_2}O \rightleftarrows {H_3}{O^ + } + O{H^ - }\]
The equilibrium constant is given by, \[{{\text{K}}_{{\text{eq}}}}\]
\[{{\text{K}}_{{\text{eq}}}}{\text{ = }}\dfrac{{\left[ {{{\text{H}}^{\text{ + }}}} \right]\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]}}{{{\text{[}}{{\text{H}}_{\text{2}}}{\text{O]}}}}\]
And the ionic product of water is given by,
\[{{\text{K}}_{\text{w}}}{\text{ = }}\left[ {{{\text{H}}^{\text{ + }}}} \right]\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]\]
Or
\[{{\text{K}}_{\text{w}}}{\text{ = }}\left[ {{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}} \right]\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]\]
And,
\[{{\text{K}}_{\text{w}}}{\text{ = }}{{\text{K}}_{{\text{eq}}}}\left[ {{{\text{H}}_{\text{2}}}{\text{O}}} \right]\]
We know that,
\[{{\text{K}}_{{\text{eq}}}}{\text{ = }}\dfrac{{{{\text{K}}_{\text{w}}}}}{{{\text{55}}{\text{.4M}}}}\]
=\[1.1 \times {10^{ - 16}} \times 55.4\]

Thus, the correct option is C.

Additional information:
With increase in temperature, the ionic product of water also increases. As the temperature increases, the degree of ionization also increases. And there is more dissociation of water into \[{{\text{H}}^{\text{ + }}}\]and \[{\text{O}}{{\text{H}}^{\text{ - }}}\] ions. This results in the increase in the concentration of these two ions. And hence the ionic product also increases.
In case of an acid or a base, the increase of decrease in the concentration of \[{\text{H}}{{\text{O}}_{{{\text{3}}^{}}}}^{\text{ + }}\]can be explained with the help of Le Chatlier Principle.
\[{\text{2}}{{\text{H}}_{\text{2}}}{\text{O}} \rightleftarrows {{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}{\text{ + O}}{{\text{H}}^{\text{ - }}}\]
On addition of some pure water to acid, the \[{\text{H}}{{\text{O}}_{\text{3}}}^{\text{ + }}\] concentration increases and the equilibrium shifts in the backward direction.
If base is added \[{\text{O}}{{\text{H}}^{\text{ - }}}\] concentration increases and in this case also, the equilibrium shifts in the backward direction.

Note:
Let us note that an ionisation constant is a value that depends on the equilibrium between ions and non-ionized molecules in a solution or liquid. It's the ratio of products to reactants increased to the right stoichiometric powers, or the ratio of concentration product to reactant. An ionization constant is also known as a dissociation constant. Ionization constant quantifies the degree a substance will ionize in a solution (typically water).