Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0163 M solution. The pH of the resulting solution is 6.41. Calculate the pKa for the acid.

Answer 1

PH = 6.41

-log[H+] = 6.41

[H+]      = 10^-6.41 = 3.9*10^-7

         HA -----> H+ + A-

I       C           0       0

C     -C      C      C

E    C(1-)   C      C

ka   = [H+][A-]/[HA]

      = C * C/C(1-)

Ka = C²

=Ka/C

[H+] = CKa/C

[H+] = KaC

3.9*10^-7      = KaC

(3.9*10^-7)²   = KaC

15.21*10^-14   = Ka*0.0163

Ka   = 9.33*10^-12