Calculate the pH of 0.215 M carbonic acid (H2CO3). (HINT: do NOT worry about the second H+ in this acid!)
How would I solve this?
Carbonic acid is a weak acid.
H2CO3 + H2O---> HCO3 +H30+
Initial H2CO3=0.215 M
Final H2CO3=0.215-x
Initial HCO3=0
Final HCO3= x
Initial H30+=0
Final H30+= x
ka=4.3 x 10-7=[H30+][HCO3]/H2CO3]
4.3 x 10-7=[x][x]/0.215M-x]
5% rule allows for ignoring -x
4.3 x 10-7=[x][x]/0.215M]
sqrt(4.3 x 10-7 *0.215M])=x
x=H3O+
pH=-log[H3O+]
To solve this problem, you need to use the concept of equilibrium and the acid dissociation constant (Ka) of carbonic acid (H2CO3).
First, write the balanced chemical equation for the dissociation of carbonic acid:
H2CO3 ⇌ H+ + HCO3-
The Ka expression for this reaction would be:
Ka = [H+][HCO3-] / [H2CO3]
Since we are instructed to not worry about the second H+ in this acid, we can assume that the concentration of H+ is equal to the concentration of H2CO3 because the dissociation reaction goes almost to completion.
Therefore, we have:
Ka = [H+]^2 / [H2CO3]
Now, you need to know the value of Ka for carbonic acid. The value of Ka for carbonic acid is 4.45 x 10^-7.
Next, you substitute the given concentration of carbonic acid (0.215 M) into the Ka expression and solve for the concentration of H+ (pH):
4.45 x 10^-7 = [H+]^2 / 0.215
Solving for [H+], we get:
[H+]^2 = 4.45 x 10^-7 * 0.215
[H+]^2 = 9.5585 x 10^-8
[H+] = sqrt(9.5585 x 10^-8)
[H+] = 3.091 x 10^-4 M
Finally, to find the pH, you take the negative logarithm (base 10) of the H+ concentration:
pH = -log([H+])
pH = -log(3.091 x 10^-4)
pH ≈ 3.51
Therefore, the pH of a 0.215 M solution of carbonic acid is approximately 3.51.