butanoic acid PKa= 4.82 A buffer solution is prepared by mixing 82.3 ml of 0.032
ID: 1066416 • Letter: B
Question
butanoic acid PKa=4.82
A buffer solution is prepared by mixing 82.3 ml of 0.0329 M butanoic add with 21.4 mL of 0.661 M sodium butanoate. A table of pKa values can be found here. Calculate the pH (to two decimal places) of this solution. Assume the 5% approximation is valid and that the volumes are additive. Calculate the pH (to two decimal places) of the buffer solution after the addition of 13.9 mL of a 0.0525 M solution of hydrochloric acid to the existing buffer solution. Assume the 5% approximation is valid and that the volumes are additive.Explanation / Answer
1) [butanoic acid] = molarity x volume in Litres = 0.0329 M x 0.0823 L = 0.0027 mol
[sodium butanoate] = molarity x volume in Litres = 0.661 M x 0.0214 L = 0.014 mol
pKa of butanoic acid = 4.82
According to Henderson-Hasselbalch equation,
pH = pKa + log [sodium butanoate]]/ [butanoic acid ]
= 4.82 + log ( 0.0414 mol/ 0.0027 mol)
= 6.00
pH = 6.00
2) pH after addition of HCl
[HCl] = molarity x volume in Litres = 0.0525 M x 0.0214 L = 0.0139 mol
sodium butanoate + HCl ------------> butanoic acid + NaCl
0.014 mol 0.0139 mol 0
----------------------------------------------------------------------------------------------------
0.014 -0.0139 0 0.0139 mol
= 0.0001 mol
Hence,
[sodium butanoate] = 0.0001 mol
[butanoic acid] = 0.027 mol + 0.0139 mol = 0.0409 mol
Then,
pH = pKa + log [sodium butanoate]]/ [butanoic acid ]
= 4.82 + log ( 0.0001 mol/ 0.0409 mol)
= 2.2
Therefore,
pH after addition of HCl = 2.2
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