Question

The ionization constant and degree of dissociation of water at ${25}^{o}C$, is:

• A. $1.8\times {10}^{-9},1.8\times {10}^{-16}$
• B. $1.8\times {10}^{-16},1.8\times {10}^{-9}$
• C. $1.8\times {10}^{-12},1.8\times {10}^{-14}$
• D. $1.8\times {10}^{-14},1.8\times {10}^{-12}$

Answer 1

Answer: (A)

$1.8\times {10}^{-9},1.8\times {10}^{-16}$

Degree of dissociation(h)$=\frac{K_w}{C}$

$K_w=H^{+}\times O{H}^{-}={10}^{-14}$ and $C=55.5$mol/L

Degree of dissociation$=\frac{{10}^{-14}}{55.5}=1.8\times {10}^{-16}$

Ionization constant $=\frac{h}{H^{+}}=\frac{1.8\times {10}^{-16}}{{10}^{-7}}=1.8\times {10}^{-9}$