posted by Meenaakshi on .
Given acidity constant,Ka of benzoic acid is 6.28 X 10^-5. pH of 0.15 molar solution of this acid is ?
Here is my method:
Benzoic acid is a weak acid,hence it dissociates very little. So:
C6H5COOH---> C6H5COO- + H+
[H+] and [C6H5COO-] are yet to be calculated,so let them be x
[C6H5COOH] = 0.15-x
Since Ka= 6.28 X 10^-5 mol/L,using the Kc equation:
Ka = x.x/(0.15-x)
Assuming x<< 0.15,simplifying the eqn becomes : Ka = x.x/(0.15)
Solving the equation,x value is 3.069201851 X 10^-3 mol/L = [H+].
pH = -log[H+] = 2.51. Is this correct?
I agree. Thanks for showing your work. That helps a lot.