What is the pH of a 2.23 M HCHO2, formic acid, aqueous solution at 25 °C? ( Ka = 1.8E-4 )
............HCOOH ==> H^+ + HCOO^-
begin....... 2.23M....0.......0
change.......-x.......x........x
final......2.23-x.....x........x
Ka = (H^+)(HCOO^-(/(HCOOH)
Substitute from the ICE chart into Ka expression and solve for (H^+). Convert to pH.
To determine the pH of the solution, you first need to find the concentration of H+ ions by calculating the value of Ka, which is the acid dissociation constant. The equation for the equilibrium constant expression is:
Ka = [H+][CHO2-] / [HCHO2]
Given that [CHO2-] is equal to the concentration of HCHO2 (since one mole of HCHO2 will yield one mole of H+ and one mole of CHO2-), you can rewrite the equation as:
Ka = [H+]^2 / [HCHO2]
Rearranging the equation gives:
[H+]^2 = Ka x [HCHO2]
[H+] = √(Ka x [HCHO2])
Now, substitute the given values into the equation.
Ka = 1.8E-4
[HCHO2] = 2.23 M
[H+] = √(1.8E-4 x 2.23)
= √(0.0003234)
≈ 0.018
Finally, calculate the pH by taking the negative logarithm (base 10) of the concentration of H+ ions:
pH = -log10([H+])
= -log10(0.018)
≈ 1.75
Therefore, the pH of a 2.23 M HCHO2 solution is approximately 1.75 at 25 °C.