An analytical chemist is titrating 160.3mL of a 1.100M solution of acetic acid HCH3CO2 with a 0.4100M solution of NaOH. The pKa of acetic acid is 4.70. Calculate the pH of the acid solution after the chemist has added 114.1mL of the NaOH solution to it.
Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added.
Round your answer to 2 decimal places.
Given:
M(HCH3CO2) = 1.1 M
V(HCH3CO2) = 160.3 mL
M(NaOH) = 0.41 M
V(NaOH) = 114.1 mL
mol(HCH3CO2) = M(HCH3CO2) * V(HCH3CO2)
mol(HCH3CO2) = 1.1 M * 160.3 mL = 176.33 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.41 M * 114.1 mL = 46.781 mmol
We have:
mol(HCH3CO2) = 176.33 mmol
mol(NaOH) = 46.781 mmol
46.781 mmol of both will react
excess HCH3CO2 remaining = 129.549 mmol
Volume of Solution = 160.3 + 114.1 = 274.4 mL
[HCH3CO2] = 129.549 mmol/274.4 mL = 0.4721M
[CH3CO2-] = 46.781/274.4 = 0.1705M
They form acidic buffer
acid is HCH3CO2
conjugate base is CH3CO2-
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.7+ log {0.1705/0.4721}
= 4.258
Answer: 4.26
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