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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