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You prepare a buffer solution by dissolving 2.00 g each of benzoic acid, C6H5COOH (mm = 122 g/mol, ka = 6.3 x 10-5 ) and sodium benzoate, NaC6H5COO (mm = 144 g/mol) in 750.0 ml of water. What is the ph of this buffer? Assume that the solution's volume is 750.0 ml.rnrnenter your answer to 2 decimal places.
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The Henderson-Hasselbalch equation can be used to calculate the pH of a buffer solution:
pH = pKa + log([A-]/[HA])
where pKa is the negative logarithm of the acid dissociation constant, [A-] is the molar concentration of the conjugate base, and [HA] is the molar concentration of the acid.
In this case, the pKa of benzoic acid is 6.3 x 10-5. The molar concentration of benzoic acid and sodium benzoate is:
[C6H5COOH] = 2.00 g / (122 g/mol) / 0.750 L = 0.0260 M
[NaC6H5COO] = 2.00 g / (144 g/mol) / 0.750 L = 0.0218 M
Substituting these values into the Henderson-Hasselbalch equation gives:
pH = 4.20 + log(0.0218/0.0260) = 4.18
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