use:
pKa = -log Ka
4.75 = -log Ka
Ka = 1.778*10^-5
Given:
M(CH3COOH) = 0.25 M
V(CH3COOH) = 100 mL
M(NaOH) = 1.25 M
V(NaOH) = 10 mL
mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.25 M * 100 mL = 25 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 1.25 M * 10 mL = 12.5 mmol
We have:
mol(CH3COOH) = 25 mmol
mol(NaOH) = 12.5 mmol
12.5 mmol of both will react
excess CH3COOH remaining = 12.5 mmol
Volume of Solution = 100 + 10 = 110 mL
[CH3COOH] = 12.5 mmol/110 mL = 0.1136M
[CH3COO-] = 12.5/110 = 0.1136M
They form acidic buffer
acid is CH3COOH
conjugate base is CH3COO-
Ka = 1.778*10^-5
pKa = - log (Ka)
= - log(1.778*10^-5)
= 4.75
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.75+ log {0.1136/0.1136}
= 4.75
Answer: 4.75
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