Question

# Consider the titration of a 60.0 mL of a 0.20M formic acid solution (HCHO2) with 0.10M...

Consider the titration of a 60.0 mL of a 0.20M formic acid solution (HCHO2) with 0.10M NaOH. A) Calculate the initial pH of the formic acid solution, before any base is added. B) At what point in the titration is pH = pKa? What volume of the NaOH will have been added to get to this point? C) Calculate the pH after 30.0 mL of the base has been added.

formic acid is weak acid with pKa 3.75

A) for weak acids pH = 1/2 [pKa - log C]

pH = 1/2 [3.75 - log 0.2]

pH = 1/2 [4.45]

pH = 2.22

B) when we titrate weak acid with NaOH following reaction takes place.

HCHO2 + NaOH ----------> NaCHO2 + H2O

mixture of NaCHO2 and HCHO2 act as acidic buffer.

for acidic buffer

pH = pKa + log [NaCHO2] / [HCHO2]

at equivalence point [NaCHO2] = [HCHO2]

so log [NaCHO2]/[HCHO2] = 0

then pH = pKa

millimoles of acid = 60 x 0.2 = 12

12 / 2 = 6.0 millimoles NAOH must be added to reach half equivalence point.

6 = V x 0.1

V = 60 mL NaOH must be added.

C) millimoles of NaOH added = 30 x 0.1 = 3.0

12 - 3 = 9 millimoles acid left

3 millimoles salt will form.

[salt] = 3 / 90 = 0.033 M (90 ml is total volume acid + base)

[acid] = 9 / 90 = 0.1 M

pH = pKa + log [salt] /[acid]

pH = 3.75 + log [0.033]/[0.1]

pH = 3.75 - 0.48

pH = 3.27

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