Chemistry · General chemistry II · Worked example
Add a strong base to a buffer
A 1.00 L buffer contains 0.100 mol acetic acid and 0.100 mol sodium acetate. Find its pH after adding 0.010 mol NaOH, and compare with adding the same base to 1.00 L of water.
The starting pH
Equal amounts of acid and base: pH = pKa = 4.74.
React the base first
OH⁻ converts acetic acid into acetate: HC₂H₃O₂ + OH⁻ → C₂H₃O₂⁻ + H₂O. The acid falls by 0.010 mol and the base rises by 0.010 mol.
Apply Henderson–Hasselbalch
Use the new amounts; the volume cancels in the ratio.
Compare with water
In 1.00 L of water the same NaOH gives [OH⁻] = 0.010 M, pOH 2.00 and pH 12.00: a jump of five units, against 0.09 for the buffer.
Result
pH 4.83, up only 0.09; in water the pH would jump from 7.00 to 12.00.
Your turn
Find the pH after adding 0.015 mol HCl to the original buffer instead.
Show the answer and explanation
pH = 4.61.
H₃O⁺ turns acetate into acetic acid: 0.100 − 0.015 = 0.085 mol acetate and 0.100 + 0.015 = 0.115 mol acid. pH = 4.74 + log(0.085/0.115) = 4.74 − 0.13 = 4.61.
Keep exploring
In Buffers & titration curves, add 0.050 mol NaOH instead, five times as much. The buffer still holds the pH at 5.22.
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