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An analytical chemist is titrating 71.3 mL of a 0.5300 M solution of formic acid (H2CO2) with a 0.5600 M solution of KOH. The

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Answer #1

use:

pKa = -log Ka

3.74 = -log Ka

Ka = 1.82*10^-4

Given:

M(HCOOH) = 0.53 M

V(HCOOH) = 71.3 mL

M(KOH) = 0.56 M

V(KOH) = 45.2 mL

mol(HCOOH) = M(HCOOH) * V(HCOOH)

mol(HCOOH) = 0.53 M * 71.3 mL = 37.789 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.56 M * 45.2 mL = 25.312 mmol

We have:

mol(HCOOH) = 37.789 mmol

mol(KOH) = 25.312 mmol

25.312 mmol of both will react

excess HCOOH remaining = 12.477 mmol

Volume of Solution = 71.3 + 45.2 = 116.5 mL

[HCOOH] = 12.477 mmol/116.5 mL = 0.1071M

[HCOO-] = 25.312/116.5 = 0.2173M

They form acidic buffer

acid is HCOOH

conjugate base is HCOO-

Ka = 1.82*10^-4

pKa = - log (Ka)

= - log(1.82*10^-4)

= 3.74

use:

pH = pKa + log {[conjugate base]/[acid]}

= 3.74+ log {0.2173/0.1071}

= 4.047

Answer: 4.05

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