Question

The degree of dissociation of water at ${25}^{\circ }C$ is $1.9\times {10}^{-7}$ and density is $1.0gc{m}^{-3}$ 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. $1.0\times {10}^{-8}$

Answer 1

The correct option is B $1.0\times {10}^{-14}$

$H_{2}O\leftrightharpoons H_{ac}^{+}+O{H}_{ac}^{-}\alpha =\frac{1.9\times {10}^{-7}}{100}$ ${\rho }_{water}=1.0gc{m}^{-3}\therefore c=\frac{1}{18}\times 1000=55.56moles\therefore \left[H^{+}\right]=55.56\times 1.9\times {10}^{-9}=1.055\times {10}^{-7}\therefore K_{\Psi }=\left[H^{+}\right]\left[O{H}^{-}\right]=(1.055\times {10}^{-7}{)}^2=1.0\times {10}^{-14}$