Question

The atmospheric sulfur dioxide concentration over a certain region is 0.12 ppm by volume. Calculate the PH of the rainwater due to this pollutant. Assume that the dissolution of does not affect its pressure.

Answer 1

To determine the pH, we need to find the concentration of hydrogen and hydroxide ions in a solution. It can be calculated with the following formula.

`pH=-\log \lbrack H^+\rbrack`

Givens.

• The concentration of Sulfur Dioxide in the atmosphere is 0.12 ppm.

The chemical equation about the situation is

`SO_2+H_2O\to H+HSO_3`

We can the hydrogen ions products whose concentration we need to find.

We know that 1 ppm is equivalent to 1x10^-6 atm. Let's find the pressure of SO2 in atm.

`0.12\text{ppm}\cdot\frac{1*10^(-6)\text{atm}}{1\text{ppm}}=1.2*10^(-7)atm`

Then, determine the equilibrium concentration Ka.

`K_a=(\lbrack H\rbrack\lbrack HSO_3\rbrack)/(\lbrack P_(SO2)\rbrack)`

As you can observe, the pressure in the atmosphere is the concentration of SO2. Also, we know that the constant of equilibrium of HSO3 is 1.3x10^-2, so

`1.3*10^(-2)=(x\cdot x)/(1.2*10^(-7))`

Then, solve for x.

`\begin{gathered} x^2=1.3*10^(-2)\cdot1.2*10^(-7) \\ x=\sqrt[]{1.56*10^(-9)_{}} \\ x=3*10^(-5) \end{gathered}`

Where x represents the concentration of hydrogen ions and HSO3. Now, we are able to find the pH.

`pH=-\log \lbrack3*10^(-5)\rbrack\approx4.5`

Therefore, the pH of the rainwater due to this pollutant is 4.5.