Answer:
First, we need to set up the equation for the ionization of HCHO2:
HCHO2 + H2O ↔ H3O+ + CHO2-
The Ka expression for this reaction is:
Ka = [H3O+][CHO2-]/[HCHO2]
We know the concentration of HCHO2 is 0.015 M, and the Ka value is 1.8 × 10^-4. We can use an ICE (initial, change, equilibrium) table to find the concentration of H3O+ and CHO2- at equilibrium:
HCHO2 + H2O ↔ H3O+ + CHO2-
I: 0.015 M 0 M 0 M 0 M
C: -x +x +x +x
E: 0.015-x x x x
Using the Ka expression, we can plug in our equilibrium concentrations (in terms of x):
1.8 × 10^-4 = x^2/(0.015-x)
Simplifying:
x^2 = 1.8 × 10^-4 (0.015-x)
x^2 = 2.7 × 10^-6 - 1.8 × 10^-4 x
Rearranging and using the quadratic formula:
x = [1.8 × 10^-4 ± sqrt((1.8 × 10^-4)^2 - 4(1)(-2.7 × 10^-6))] / 2(1)
x = 0.0136 or 0.00108
We reject the 0.0136 value, since it is greater than our initial concentration of 0.015 M. Therefore, our equilibrium concentration of H3O+ is 0.00108 M.
To find the pH, we take the negative logarithm of the H3O+ concentration:
pH = -log(0.00108) = 2.97
Therefore, the pH of a 0.015 M solution of HCHO2 is 2.97.