Calculate the pH of a solution prepared by adding 36 cm3 of 0.15 M methanoic acid to 27 cm3 of 0.20 M NaOH.�(Ka for methanoic acid = 1.77 x 10-4)
See your other post.
mmols HCOOH = mL x M = estimated 5.4
mmols NaOH = mL x M = estd 5.4
......HCOOH + NaOH ==> HCOONa + H2O
I.......5.4.....0........0.......0
add............5.4...............
C.......-5.4..-5.4.....5.4........
E.......0.......0.......5.4
So this is not a buffer problem because you have no acid/base mixture. The base is there (5.4 mmols HCOONa) but the acid was neutralized; therefore, the pH is determined by the hydrolysis of the salt (HCOONa)
(HCOONa) = 5.4 millimols/63 mL volume = 0.086M
........HCOO^- + HOH ==> HCOOH + OH^-
I.......0.086..............0......0
C.........-x...............x......x
E......0.086-x.............x......x
Kb for HCOONa = (Kw/Ka for HCOOH) = (x)(x)/(0.086-x)
Solve for x = OH^- and convert to pH.