Carbonic acid forms when carbon dioxide dissolves in water. A typical can of soda contains a 0.120 M solution of CO2. Assuming that all dissolved CO2 is present as carbonic acid, and that no other substances dissolved in solution affect the pH, what is the pH of this solution?
H2CO3: Ka1 = 4.68 x 10-7, Ka2 = 5.6*10-11
I'm not sure how to get from the concentration of CO2 to the Concentration of the H2CO3 reaction to get the pH
I think the secret is "assuming that all dissolved CO2 is present as H2CO3".
.......H2CO3 ==> H^+ + HCO3^-
I.......0.12......0.....0
C.......-x........x.....x
E......0.12-x.....x.....x
Set up the E line into k1 and solve for H^+ and convert to pH.
To determine the pH of the carbonic acid solution, you can use the information given about the dissociation constants (Ka) of carbonic acid.
First, it's important to understand that the dissolved CO2 actually exists in equilibrium with carbonic acid (H2CO3) in water. This equilibrium can be represented by the following equation:
CO2 + H2O ⇌ H2CO3
Since the concentration of CO2 in the soda is given as 0.120 M, we can assume that the initial concentration of carbonic acid (H2CO3) is also 0.120 M.
The dissociation of carbonic acid occurs in two stages:
1. H2CO3 ⇌ H+ + HCO3-
2. HCO3- ⇌ H+ + CO3^2-
The values of the Ka1 and Ka2 given (4.68 x 10^-7 and 5.6 x 10^-11, respectively) are the acid dissociation constants for these reactions.
To find the equilibrium concentrations of H2CO3, HCO3-, and CO3^2-, you will need to use an ICE (Initial, Change, Equilibrium) table for each dissociation step.
For the first dissociation step:
H2CO3 ⇌ H+ + HCO3-
Initial concentration: [H2CO3] = 0.120 M
Change in concentration: -x +x +x (since one molecule dissociates into one H+ and one HCO3-)
Equilibrium concentration: [H2CO3] - x, [H+] = x, [HCO3-] = x
For the second dissociation step:
HCO3- ⇌ H+ + CO3^2-
Initial concentration: [HCO3-] = x (from the first step)
Change in concentration: -x +x +x (again, one molecule dissociates into one H+ and one CO3^2-)
Equilibrium concentration: [HCO3-] - x, [H+] = x, [CO3^2-] = x
Now, you can write the expressions for the Ka1 and Ka2 equilibrium constants:
Ka1 = [H+][HCO3-] / [H2CO3]
Ka2 = [H+][CO3^2-] / [HCO3-]
Substituting the equilibrium concentrations for each species, you get:
Ka1 = x^2 / (0.120 - x)
Ka2 = x^2 / (x)
Since x is typically small compared to the initial concentration of H2CO3 (0.120 M), we can approximate (0.120 - x) as 0.120 in the denominator of the Ka1 expression.
Now, for both Ka1 and Ka2, solve for x (the concentration of H+), using the given Ka values.
Using the quadratic equation (because these are quadratic equations), you can solve for x. The positive root should be selected as the concentration of H+.
Once you have the concentration of H+, you can calculate the pH using the equation:
pH = -log10 [H+]
Remember to keep in mind the assumption made in the problem that no other substances dissolved in solution affect the pH. In real-life situations, soda contains other acidic or basic components that can influence the final pH value.