A buffer solution is prepared by mixing 82.9 mL of 0.934 M butanoic acid with 70
ID: 1001747 • Letter: A
Question
A buffer solution is prepared by mixing 82.9 mL of 0.934 M butanoic acid with 70.6 mL of 0.579 M sodium butanoate. A table of pKa values can be found here. 1. Calculate the pH (to two decimal places) of this solution. Assume the 5% approximation is valid and that the volumes are additive.
Ans. Part 1 pH calculated to be 4.54.
2. Calculate the pH (to two decimal places) of the buffer solution after the addition of 3.35 g of sodium butanoate (NaC3H7COO) to the buffer solution above. Assume 5% approximation is valid and that the volume of solution does not change.
Explanation / Answer
The pH of buffer solution can be caluclated using Hendersen Hassalbalch equation, which is
pH = pKa + log[salt] / [acid]
pKa of butanoic acid = 4.82
Moles of butanoic acid = Molarity X volume = 82.9 X 0.934 = 77.43 millimoles
Moles of salt of acid = 70.6 X 0.579 = 40.88 millimoles
pH = 4.82 + log [40.88] / [77.43] = 4.54 (Answer)
2) Now we have added 3.35 g of sodium butanoate
Moles = Mass / mol wt = 3.35 / 110 = 0.0305 =
New pH will be
pH = pKa + log [Salt] / [acid]
Moles of salt = 30.5 + 40.88 = 71.38
pH = 4.82 + log [71.38 / 77.43]
pH = 4.82 + (-0.03) = 4.78 (approx)
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