Question

At ${80}^{\circ }C$, distilled water has hydronium ion $\left(H_3{O}^{+}\right)$ concentration equal to $1\times {10}^{-6}mol/L$. The value of $K_w$ at this temperature would be:

• A. $1\times {10}^{-6}$
• B. $1\times {10}^{-12}$
• C. $1\times {10}^{-9}$
• D. $1\times {10}^{-14}$

Answer 1

The correct option is B $1\times {10}^{-12}$

The distilled water is neutral with equal concentrations of hydrogen ion and hydroxide ion. Whatever the tempearture may be water will always dissociate to give equal conc. of $H^{+}$ ion and $O{H}^{-}$ ion, that is it will always be neutral.
Hence, $K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$
=$1\times {10}^{-6}\times 1\times {10}^{-6}$
=$1\times {10}^{-12}$

Hence, the correct option is B.



Theory :
Ionic Product of water:
$H_{2}O\left(l\right)+H_{2}O\rightleftharpoons H_3{O}^{+}+O{H}^{-}$ or $H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$ as per laws of mass action : $K_c=\frac{\left[H^{+}\right]\left[O{H}^{-}\right]}{\left[H_{2}O\right]}$
In pure water and in dilute aqueous solutions. The molar concentration of water is essentially a constant, with molarity of $55.55mol/L$.
Density of $H_{2}O=1g/mL$ or $H_{2}Oing=Volumeof{H}_{2}OinmL=1gor1mL$
$massof{H}_{2}Oin1Lofwater=1000g$
$Molarity\left(M\right)=\frac{molesofwater}{Volumeofsolution}$ $M=\frac{\frac{1000}{18}}{1}=55.55mol/L$
The ratio of dissociated water to that of undissociated water can be given as :
$\frac{{10}^{-7}}{55}=1.8\times {10}^{-9}$. Thus, equilibrium lies mainly towards undissociated water.
Therefore, $K_c\times \left[H_{2}O\right]=\left[H^{+}\right]\left[O{H}^{-}\right]$ $K_c\times H_{2}O$ is a constant which is also called ionic product of water $K_w$.
For pure water at ${25}^{o}C$ $\left[H^{+}\right]=\left[O{H}^{-}\right]={10}^{-7}M$ and $K_w=\left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-14}{M}^2$

Effect of temperature on $K_w$:
The ionic product of water $K_w$ increases with increase of temperature. This is because of the fact that with increase of temperature, the degree of ionization of water increases.
$K_w$ is proportional to temperature.
$K_w\propto T$
Note:- Ionic product of water is always a constant whatever may be dissolved in water. As it is an equilibrium constant, it will depend only on temperature.