Question

Calculate the pka for the weak acid HA that is 2.3% ionized in a .080M solution. How would I do this?

Answer 1

for the pKa value first let us find percent ionisation to molarity

As 100% gives 0.08M concentration of H+ or pH of solution is -log(0.08) =1.096

2.3% ionisation = 2.3*0.08/100=1.84*10^-3M concentration of H⁺

K_a = [A^-]_eq [H⁺]_eq/[HA]_eq

We know the equilibrium concentration of H⁺, since we were given the pH of the solution.  [H⁺]_eq = 1.84*10^-3M

2.  We know the initial concentration of HA.  We will lose x moles of this as the acid undergoes dissociation to form A^- and H⁺.

3.                                 [HA](M)                 [H⁺](M)           [A^-](M)

            Initial                0.08 0.00                 0.00

            Change             -x                                 +x                    +x

            Equilibrium       0.08 -x x                      x

  K_a   =   (x)(x)

                      (0.08 - x)

4.  Assume that all of the H⁺ comes from the acid, and none from water.   We know that the [H⁺]_eq = 1.84*10^-3M

We can also see that the concentration of HA will change very little, from 0.08 to 0.08 - 1.84 x 10^-3.  The change in concentration can be ignored if it is less than 5% of the original concentration.  0.08 M x 5%, so the change in [HA] in this problem can be ignored.

Ka=(1.84*10^-3)²/0.08=4.2*10^-5

Ka = 10^(-pKa)

so pKa=4.37