[Sis290025]

Calculate the pH of a 0.124 M sodium formate solution (HCOONa). K_a of HCOOH is 5.9*10^-2.

HCOONa --> HCOO- + Na+

HCOO- + H2O --> HCOOH + OH-

K_b = K_w/K_a = (1.0*10^-14)/(5.9*10^-2) = 1.6949152E-13

K_b = [OH-][HCOOH]/[HCOO-]

1.6949152E-13 = x^2/[0.124 -x]

Assuming that 0.124 - x = 0.124.

x = sqrt(0.124*K_b) = 1.449722E-7

[OH-] = 1.449722E-7 M

pOH- = -log(1.449722E-7) = 6.838715

pH = 14 - pOH- = 14 - 6.838715 = 7.16

[Alberto_Kravina]

Correct!

I calculated it like this:

pKa = - lg(Ka)

pKb = 14 - pKa

pOH = 0.5(pKb-lg(c₀))

pH = 14 - pH

[Borek]

Ka of HCOOH is 10^-3.75, but it doesn't matter.

Your approach is correct and the result should be accepted by your teacher.

Don't read further

Final result is slightly off. You did assumption that [HCOOH] = [OH^- ]. Exact calculations (done numerically) show that it is not a case:

[HCOOH] = 1.20*10^-7
[OH^- ] = 1.75*10^-7

and exact pH = 7.24

Edit: Alberto was faster

[Alberto_Kravina]

Edit: Alberto was faster

Yes- but you didn't notice that I made a mistake...

[qlabel]

Yes- but you didn't notice that I made a mistake...

What mistake?