Question

If the ?aKa of a monoprotic weak acid is 6.2×10−66.2×10−6, what is the pH of a 0.24 M0.24 M solution of this acid?

Enough of a monoprotic weak acid is dissolved in water to produce a 0.01130.0113 M solution. The pH of the resulting solution is 2.352.35. Calculate the Ka for the acid.

Answer 1

1)

Ka = 6.2 x 10^-6

pKa = -log Ka = -log ( 6.2 x 10^-6 )

pKa = 5.21

pH = 1/2 [pKa -log C]

pH = 1/2 [5.21 -log 0.24]

pH = 2.92

2)

[H+] = x   = 10^-2.35 = 4.67 x 10^-3 M

C = 0.0113 M

Ka   = [H+]^2 / C-x

Ka = (4.67 x 10^-3)^2 / 0.0113 - (4.67 x 10^-3)

Ka = 3.01 x 10^-3