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Question: The degree of dissociation of water at \({25^{\text{o}}}{\text{C}}\) is \(1.8 \times {10^{ - 7}}\% \...

The degree of dissociation of water at 25oC{25^{\text{o}}}{\text{C}} is 1.8×10−7%1.8 \times {10^{ - 7}}\% and density is 1.0gcm−31.0{\text{gc}}{{\text{m}}^{ - 3}}. The ionic constant for water is:
A.1.0×10−141.0 \times {10^{ - 14}}
B.2.0×10−162.0 \times {10^{ - 16}}
C.1.0×10−161.0 \times {10^{ - 16}}
D.1.0×10−81.0 \times {10^{ - 8}}

Explanation

Solution

Concentration of water is the ratio of its moles per litre. Ionization constant is the product of degree of dissociation and concentration of water. It is represented by Ka{K_a}. We shall calculate the concentration of 1 litre of water from the density and substitute it in the ionization constant formula.
Formula Used: Ka=Cα2{K_a} = C{\alpha ^2}
Where C is the concentration of water, α\alpha is the degree of dissociation and Ka{K_a} is the ionization constant.

Complete step by step answer:
Water is a weak electrolyte and thus, cannot ionize completely.
First we need to calculate the concentration of water which is known to be the moles per litre. The density of water is given as 1.0gcm−31.0{\text{gc}}{{\text{m}}^{ - 3}}.
Thus, the mass of one litre of water which is 1000cm - 31000{\text{c}}{{\text{m}}^{{\text{ - 3}}}} is equal to 1000g1000{\text{g}}. The molar mass of water is known as 18g18{\text{g}}.
We know that the concentration of water is given as 100018=55.5molL - 1\dfrac{{1000}}{{18}} = 55.5{\text{mol}}{{\text{L}}^{{\text{ - 1}}}}
Water dissociates and gives hydrogen ion and hydroxyl anion. The ionic product is(Kw)=[H+][OH−]=C2α2\left( {{K_w}} \right) = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] = {C^2}{\alpha ^2}
The ionic constant is given by Ka=[H+][OH−][H2O]=C2α2C{K_a} = \dfrac{{\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ {{H_2}O} \right]}} = \dfrac{{{C^2}{\alpha ^2}}}{C}
Thus we get, Cα2=(55.56)×(1.8×10−9)2≈2×10−16C{\alpha ^2} = \left( {55.56} \right) \times {\left( {1.8 \times {{10}^{ - 9}}} \right)^2} \approx 2 \times {10^{ - 16}}

Therefore, the correct answer is B.

Note:
When the product of hydrogen ion concentration and hydroxide ion concentration is constant when temperature is kept constant is called as the ionic product of water. Water undergoes self- ionization forming hydronium ions and hydroxide ions. When two molecules of water collide in the liquid state, hydrogen ions are transferred from one molecule to the other. While writing the expression of an ionic product, concentration of water is not included because it is a pure liquid and its concentration can be considered equal to unity. At room temperature the ionic product of water is generally equal to 10−14{10^{ - 14}}.The unit of ionic product is mol2L - 2{\text{mo}}{{\text{l}}^{\text{2}}}{{\text{L}}^{{\text{ - 2}}}}.