the ph of a .030M C2H5 COOH solution is 3.20. What is this acids Ka?
Convert pH to H^+).
pH = -log(H^+)
3.20 = -log(H^+)
I get 6.3E-4 but you should confirm that.
.....C2H5COOH ==> H^+ + C2H5COO^-
I......0.030.......0.......0
C.......-6.3E-4...6.3E-4..6.3E-4
E......0.0294.....6.3E-4..6.3E-4
Ka = (H^+)(C2H5COO-)/(C2H5COOH)
Substitute from the ice chart and solve for Ka. Check all of those numbers to make sure I punched the right numbers.
To find the acid dissociation constant (Ka) of the acid, you need to use the pH and concentration of the acid solution. The pH of 3.20 indicates that the concentration of H+ ions in the solution is 10^(-3.20) moles per liter.
To calculate the Ka, you need to assume the acid fully dissociates into H+ ions and the conjugate base. In this case, the dissociation reaction is as follows:
C2H5COOH ⇌ C2H5COO- + H+
Since the concentration of C2H5COOH initially is 0.030 M, and assuming it fully dissociates, the concentration of H+ ions and C2H5COO- ions will be the same.
Let's call the concentration of H+ ions, [H+].
[H+] = [C2H5COO-] = 10^(-pH) = 10^(-3.20) M
Now, substitute this value into the equilibrium expression for Ka:
Ka = [C2H5COO-][H+] / [C2H5COOH]
Since [C2H5COO-] = [H+], the equation simplifies to:
Ka = ([H+])^2 / [C2H5COOH] = (10^(-3.20))^2 / 0.030
Calculating this will give you the Ka of the C2H5COOH acid.