A 0.942 M sample of carbonic acid, H2CO3, has a measured hydronium ion concentration of 6.36 ´ 10–4 M. Calculate the acid-ionization constant of carbonic acid. The equilibrium equation is:
H2CO3(aq) + H2O(l) <--><--> H3O+(aq) + HCO-3(aq)
Ka = (H3O^+)(HCO3^-)(H2CO3)
Set up an ICE chart, substitute into Ka expression, and solve for Ka.
Post your work if you get stuck.
what is an ICE chart?
ICE.
I = initial
C = change
E = equilibrium.
H2CO3 + H2O ==> H3O^+ + HCO3^-
Ka = (H3O^+)(HCO3^-)/(H2CO3).
initial:
H2CO3 = 0.942 M
(H3O^+) = 0
(HCO3^-) = 0
equilibrium:
(H3O^+) = 6.36 x 10^-4 M
(HCO3^-) = 6.36 x 10^-4 M
(H2CO3) = 0.942-6.36 x 10^-4 = 0.942
This solution assumes Ka is the only ionization constant for H2CO3; in reality, H2CO3 has a k1 and a k2 but we are ignoring k2. The error is small.
To calculate the acid-ionization constant (Ka) of carbonic acid (H2CO3), we can use the equilibrium expression for the reaction:
Ka = [H3O+][HCO3-] / [H2CO3]
Given that the concentration of H2CO3 is 0.942 M and the concentration of H3O+ is 6.36 ´ 10–4 M, we need to find the concentration of HCO3-.
Since the reaction involves a weak acid (H2CO3) and its conjugate base (HCO3-), we can assume that the concentration of H2O remains constant. Therefore, we can treat it as a constant and ignore it in the equilibrium expression.
Let's assume the concentration of HCO3- is x M. Then, according to the stoichiometry of the reaction, the concentration of H3O+ will also be x M.
Substituting these values into the equilibrium expression:
Ka = (x)(x) / (0.942)
Since the equilibrium expression involves the square of the concentration of both H3O+ and HCO3-, we can simplify it as:
Ka = x^2 / (0.942)
Now, we need to solve this equation to find the value of x, which represents the concentration of HCO3-. Since we know the concentration of H3O+ is 6.36 ´ 10–4 M, we can substitute this value to solve for x:
(6.36 ´ 10–4)^2 / (0.942) = x^2 / (0.942)
Simplifying this equation, we find:
x^2 = (6.36 ´ 10–4)^2 × (0.942)
x^2 = 4.055 × 10–7 × 0.942
x^2 = 3.825 × 10–7
Taking the square root of both sides, we get:
x = √(3.825 × 10–7)
x ≈ 6.19 × 10–4 M
Now that we've found the concentration of HCO3-, we can substitute this value back into the equilibrium expression to calculate the acid-ionization constant (Ka):
Ka = (6.19 × 10–4)^2 / (0.942)
Ka ≈ 4.01 × 10–7
Therefore, the acid-ionization constant of carbonic acid (H2CO3) is approximately 4.01 × 10–7.