25.00 mL of 0.1 M acetic acid are titrated with 0.05 M NaOH. Calculate the pH of the solution after addition of 25 mL of NaOH. Ka(CH3COOH) = 1.8 x 10-5.
Given:
M(CH3COOH) = 0.1 M
V(CH3COOH) = 25 mL
M(NaOH) = 0.05 M
V(NaOH) = 25 mL
mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.1 M * 25 mL = 2.5 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.05 M * 25 mL = 1.25 mmol
We have:
mol(CH3COOH) = 2.5 mmol
mol(NaOH) = 1.25 mmol
1.25 mmol of both will react
excess CH3COOH remaining = 1.25 mmol
Volume of Solution = 25 + 25 = 50 mL
[CH3COOH] = 1.25 mmol/50 mL = 0.025M
[CH3COO-] = 1.25/50 = 0.025M
They form acidic buffer
acid is CH3COOH
conjugate base is CH3COO-
Ka = 1.8*10^-5
pKa = - log (Ka)
= - log(1.8*10^-5)
= 4.745
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.745+ log {2.5*10^-2/2.5*10^-2}
= 4.745
Answer: 4.75
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