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

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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

Expert Solution
Step 1

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 = 5 millimole2 mL = 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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