Question
The degree of dissociation of water at $${25^ \circ }C$$ is $$1.9 \times {10^{ - 7}}\% $$ and density is $$1.0g\,c{m^{ - 3}},$$ The ionic constant for water is
A.
$$1.0 \times {10^{ - 10}}$$
B.
$$1.0 \times {10^{ - 14}}$$
C.
$$1.0 \times {10^{ - 16}}$$
D.
$$1.0 \times {10^{ - 18}}$$
Answer :
$$1.0 \times {10^{ - 14}}$$
Solution :
TIPS/Formulae :
$$\mathop {{H_2}O}\limits_{\left( {1 + \alpha } \right)c} \rightleftharpoons \mathop {{H^ + }}\limits_{\alpha c} + \mathop {O{H^ - }}\limits_{\alpha c} $$
$$\alpha = 1.9 \times {10^{ - 7}};\,{\text{Density}}\,{\text{of water}}$$ $${\text{ = }}\,\frac{{1.0\,gm}}{{c{m^3}}}$$
$$\therefore \,\,c = \frac{1}{{18}} \times 1000 = 55.56\,{\text{moles/l}}$$
$$\therefore \,\,\left[ {{H^ + }} \right] = 55.56 \times 1.9 \times {10^{ - 9}} = $$ $$1.055 \times {10^{ - 7}}$$
$$\left[ {\because \,1.9 \times {{10}^ - }\% = 1.9 \times {{10}^{ - 9}}} \right]$$
$$\therefore \,\,{K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] = $$ $${\left( {1.055 \times {{10}^{ - 7}}} \right)^2} = 1.0 \times {10^{ - 14}}$$