if the pH of rain water is 5.6 on average what is the concentration of carbonic acid in rain water?
If the pH of rain water is 5.6 on average what is the concentration of carbonic acid in rain water?
To determine the concentration of carbonic acid in rainwater, we need to understand the relationship between pH and carbonic acid concentration. Carbonic acid (H2CO3) is formed when carbon dioxide (CO2) dissolves in water. It reacts with water to form carbonic acid:
CO2 + H2O ⇌ H2CO3
Since carbonic acid is a weak acid, it partially dissociates into hydrogen ions (H+) and bicarbonate ions (HCO3-):
H2CO3 ⇌ H+ + HCO3-
The pH scale is a measure of the concentration of hydrogen ions in a solution. It is logarithmic, meaning that each unit on the scale represents a tenfold difference in hydrogen ion concentration. The pH of 5.6 indicates that the concentration of hydrogen ions in the rainwater is 10^(-5.6) moles per liter.
In the case of carbonic acid, however, only a fraction of the hydrogen ions come from the direct dissociation of carbonic acid. The dissociation of carbonic acid into hydrogen ions and bicarbonate ions is determined by the equilibrium constant (Ka) of the reaction. The larger the Ka, the greater the extent of dissociation.
Given that carbonic acid is a weak acid with a Ka value of approximately 4.2 x 10^(-7) at 25°C, we can use the equilibrium expression to calculate the concentration of carbonic acid.
Ka = [H+][HCO3-] / [H2CO3]
Since the initial concentration of carbonic acid is equal to its dissociation constant, we can assume that the concentration of H2CO3 is equal to the concentration of H+ in rainwater.
Using the formula above and substituting the known values into the equation:
4.2 x 10^(-7) = (10^(-5.6))^2 / [H2CO3]
Simplifying the equation:
[H2CO3] = (10^(-5.6))^2 / 4.2 x 10^(-7)
[H2CO3] ≈ 4.4 x 10^(-4) moles per liter
Therefore, the concentration of carbonic acid in rainwater with a pH of 5.6 is approximately 4.4 x 10^(-4) moles per liter.
...........H2CO3 ==> H^+ + HCO3^-
initial......x.......0.......0
change..........2.51E-6..2.51E-6
equil.......x-2.51E-6
K = (H^+)(HCO3^-)/(H2CO3)
Substitute and solve for x.
Watch out. The H^+ from neutral water is 1E-7 and that's very close to this value. You may not be able to neglect that.