Name: Applications of Solubility Product
Uploaded: 2019-07-19
Duration: 8 min 36 s
Description: Solubility product helps determine the solubility of sparingly soluble salts and the condition of salt precipitation. Learn more in this video.

Question

The average concentration of

{SO}_{2}

in the atmosphere over a city on a certain day is

10 ppm

, when the average temperature is

298 K

. Given that the solubility of

{SO}_{2}

in water at

298 K

is

1.3653

moles {litre}^{- 1}

and

{pK}_{a}

of

H_{2} {SO}_{3}

is

1.92

, estimate the

pH

of rain on that day.

Accepted Answer

Given that the average concentration of ${SO}_{2}$ in the atmosphere is 10 ppm. It means 10 L of ${SO}_{2}$ present in the total volume of $10^{6} L$ of the atmospheric air.

First, let us calculate the Molar concentration of ${SO}_{2}$ .

We know that one mole of any gas occupies $22.4 L$ at STP.

Now no of moles of gas which occupies $10 L$ is

= $\frac{10 \times 273}{22.4 \times 298} = 0.4089$

The molar concentration of ${SO}_{2}$

= $\frac{no of moles of {SO}_{2}}{Total volume of air} = \frac{0.4089}{10^{6}} = 4.089 \times 10^{- 7} mol L^{- 1}$

Also given that the solubility of ${SO}_{2} = 1.3653 mol L^{- 1}$

The molar concentration of ${SO}_{2}$ is very much less than the solubility. Hence, entire ${SO}_{2}$ in air will dissolve in rainwater.

If the $c {molL}^{- 1}$ of ${SO}_{2}$ dissolved is ionized then

${SO}_{2} + H_{2} O \to H_{2} {SO}_{3} ⇌ H^{+} + {HSO}_{3}^{-} c c - x x x$

Hence, $K_{a} = \frac{[H^{+}] [{HSO}_{3}^{-}]}{[H_{2} {SO}_{3}]} = \frac{x^{2}}{c - x}$

Now substitute $K_{a} = 10^{- 1.92} = 0.012$ and $c = 4.089 \times 10^{- 7} mol L^{- 1}$

Then $x$ can be obtained from the following binomial equation

$x^{2} + K_{a} x - K_{a} c = 0$

Now, $x = \frac{- K_{a} \pm \sqrt{K_{a}^{2} + 4 K_{a} c}}{2} = 4.08 \times 10^{- 7} M$

Hence, $[H^{+}]$ produced from ${SO}_{2}$ is small hence, $[H^{+}]$ from the water must be considered.

${[H^{+}]}_{total} = {[H^{+}]}_{acid} + {[H^{+}]}_{water}$

We know that $K_{W} = {[H^{+}]}_{total} [{OH}^{-}]$

${[H^{+}]}_{total} = {[H^{+}]}_{acid} + \frac{K_{W}}{{[H^{+}]}_{total}}$

$\begin{array}{l} {[H^{+}]}_{total}^{2} - {[H^{+}]}_{acid} {[H^{+}]}_{total} - K_{W} = 0 \\ {[H^{+}]}_{total} = \frac{{[H^{+}]}_{acid} + \sqrt{{[H^{+}]}_{acid}^{2} + 4 K_{W}}}{2} \end{array}$

after substituting $[H^{+}]_{acid}, K_{W}$ values

$[H^{+}]_{total} = 4.31 \times 10^{- 7} M$

Hence, $pH = - \log [H^{+}]_{total} = - \log (4.31 \times 10^{- 7}) = 6.37$

Important Questions on Ionic Equilibrium

Learn and Prepare for any exam you want !

The average concentration of SO2 in the atmosphere over a city on a certain day is 10 ppm, when the average temperature is 298 K. Given that the solubility of SO2 in water at 298 K is 1.3653 moles litre-1 and pKa of H2SO3 is 1.92, estimate the pH of rain on that day.

Important Questions on Ionic Equilibrium

Learn and Prepare for any exam you want !