What is the pH at 25°C of a solution that results from mixing
equal volumes of a 0.0500 M solution of acetic acid and
a 0.0374 M solution of sodium hydroxide? The
pKa of acetic acid is 4.74.
Express the pH to at least two decimal places.
Let the volume of CH3COOH and NaOH be 1 mL each
Given:
M(CH3COOH) = 0.05 M
V(CH3COOH) = 1 mL
M(NaOH) = 0.0374 M
V(NaOH) = 1 mL
mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.05 M * 1 mL = 0.05 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.0374 M * 1 mL = 0.0374 mmol
We have:
mol(CH3COOH) = 0.05 mmol
mol(NaOH) = 0.0374 mmol
0.0374 mmol of both will react
excess CH3COOH remaining = 0.0126 mmol
Volume of Solution = 1 + 1 = 2 mL
[CH3COOH] = 0.0126 mmol/2 mL = 0.0063M
[CH3COO-] = 0.0374/2 = 0.0187M
They form acidic buffer
acid is CH3COOH
conjugate base is CH3COO-
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.74+ log {1.87*10^-2/6.3*10^-3}
= 5.213
Answer: 5.21
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