If 1. ml of 0.50 M formic acid (HCOOH) is titrated with the strong base of 0.50 M. KOH (Ka for HCOOH = 1.8 x 10-4)a) Determine the pH for the point at which 100. mL of KOH has been added Determine the pH for the equivalnce point
If 1. ml of 0.50 M formic acid (HCOOH) is titrated with the strong base of 0.50 M. KOH
(Ka for HCOOH = 1.8 x 10-4)a)
Determine the pH for the point at which 100. mL of KOH has been added
Determine the pH for the equivalnce point
The pH for the equivalence point :
Balance chemical equation :
HCOOH(aq) + KOH(aq) HCOOK(aq) + H2O(l)
Initially 0.50 x 1 mmol 0.50 x 1 mmol 0 excess
At equivalence point 0 0 5 mmol
Concentration of HCOO-(aq) = C = = 2.5 M
Formula : pH = 0.5 [ pKw + pKa + log C] = 0.5 x [ 14 + 3.74 + log 2.5]
pH = 7 + 1.87 + 0.2 = 9.07
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