Question

The ratio of concentration of dissociated water to that of undissociated water is $x \times 10^{-10}$ for pure water at $25^\circ$C. Find the value of x.

Answer 1

x = 18

Answer 2

For the autoprotolysis of water,

$\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-$

we have the equilibrium constant $K_w = [\text{H}^+][\text{OH}^-] = 10^{-14}$ at $25^\circ$C. In pure water,

$[\text{H}^+] = [\text{OH}^-] = 10^{-7}~\text{M}.$

The concentration of water (undissociated water) is about $55.5~\text{M}$. The ratio of the concentration of dissociated water (represented by $[\text{H}^+]$) to that of undissociated water is:

$\text{Ratio} = \frac{[\text{dissociated water}]}{[\text{undissociated water}]} = \frac{10^{-7}}{55.5} \approx 1.8 \times 10^{-9}.$

Express it in the form $x \times 10^{-10}$:

$1.8 \times 10^{-9} = 18 \times 10^{-10}.$

Thus, $x = 18$.