Determine the pH during the titration of 60.4 mL of 0.442 M formic acid (Ka = 1.8×10-4) by 0.442 M KOH at the following points. (a) Before the addition of any KOH (b) After the addition of 16.0 mL of KOH (c) At the half-equivalence point (the titration midpoint)
Determine the pH during the titration of 60.4 mL of 0.442 M formic acid (Ka = 1.8×10-4) by 0.442 M KOH at the following points.
(a) Before the addition of any KOH
(b) After the addition of 16.0 mL of KOH
(c) At the half-equivalence point (the titration midpoint)
(d) At the equivalence point
(e) After the addition of 90.6 mL of KOH
The reaction of formic acid and KOH is
HCOOH + KOH HCOO-K+ +H2O ...... (1)
(acid) (salt)
Formic acid is a weak acid. So addition of small amount of KOH, it forms a acidic buffer solution.
(a) before addition of any KOH there is only HCOOH. So, pH = -log[H+]
= -log[c] c= concentration of formic acid
= degree of dissociation
ka = c2
= = = 0.02
So, pH = -log[0.442 x 0.02] = 2.05
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