The average pH of precipitation in rural areas is 5.70. Assuming that the pH
is controlled by the carbonate system, i.e., no anthropogenic acid gases are present 
in the atmosphere, calculate the amount of calcium carbonate (mg) dissolved per liter 
of rainwater. pKa1=6.33, KCO2=3.388e-02





Hint: pCO2=3.162e-04

To calculate the amount of calcium carbonate (CaCO3) dissolved per liter of rainwater, we need to use the equilibrium expression for the carbonate system, which is:

[H+][HCO3-] / [CO2] = KCO2/pCO2

Where:
[H+] is the concentration of hydrogen ions
[HCO3-] is the concentration of bicarbonate ions
[CO2] is the concentration of dissolved carbon dioxide
KCO2 is the equilibrium constant for the reaction between CO2 and H2O
pCO2 is the partial pressure of carbon dioxide in the atmosphere

First, let's convert the given pH value to the concentration of hydrogen ions ([H+]). The pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration. So,

[H+] = 10^(-pH)

[H+] = 10^(-5.70)

[H+] = 1.99 x 10^(-6) M

Next, we need to calculate the concentration of dissolved carbon dioxide ([CO2]) using the given pCO2 value. The relationship between pCO2 and [CO2] is:

pCO2 = [CO2] / (KCO2 * [H+])

[CO2] = pCO2 * KCO2 * [H+]

[CO2] = (3.162e-04) * (3.388e-02) * (1.99 x 10^(-6))

[CO2] = 2.13 x 10^(-11) M

Now we can substitute these values into the equilibrium expression and solve for the concentration of bicarbonate ions ([HCO3-]):

[H+][HCO3-] / [CO2] = KCO2/pCO2

(1.99 x 10^(-6))(HCO3-) / (2.13 x 10^(-11)) = (3.388e-02) / (3.162e-04)

(HCO3-) = (1.99 x 10^(-6)) * (3.388e-02) * (2.13 x 10^(-11)) / (3.162e-04)

(HCO3-) = 2.25 x 10^(-13) M

Finally, to calculate the amount of calcium carbonate (CaCO3) dissolved per liter of rainwater, we need to consider the molar mass of CaCO3, which is approximately 100.09 g/mol. Conversion from molarity to milligrams per liter can be done using the molar mass of CaCO3:

Amount of CaCO3 = (HCO3-) * molar mass of CaCO3

Amount of CaCO3 = (2.25 x 10^(-13) M) * (100.09 g/mol) * (1000 mg/g)

Amount of CaCO3 = 2.25 x 10^(-8) mg/L

Therefore, the amount of calcium carbonate dissolved per liter of rainwater is approximately 2.25 x 10^(-8) mg/L.