You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic
ID: 558471 • Letter: Y
Question
You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.180 M sodium benzoate. How much of each solution should be mixed to prepare this buffer?
What is the pH of 75.0 mL of a solution that is 0.047 M in weak base and 0.059 M in the conjugate weak acid (Ka = 7.2 × 10–8)?
The half-equivalence point of a titration occurs half way to the end point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. If 0.460 moles of a monoprotic weak acid (Ka = 1.0 × 10-5) is titrated with NaOH, what is the pH of the solution at the half-equivalence point?
Explanation / Answer
1)
pH = 4.00
pKa = 4.20
total volume of buffer = 100 mL
volume of benzoic acid = x mL
volume of sodium benzoate = (100 - x ) mL
pH = pKa + log [salt / acid]
4.00 = 4.20 + log [(100-x) x 0.18 / 0.10 x]
[(100-x) x 0.18 / 0.10 x] = 0.631
18 - 0.18 x = 0.0631 x
x = 74.0
volume of benzoic acid = 74.04 mL
volume of sodium benzoate = 25.96 mL
2)
pKa = 7.14
pH = pKa + log [base/ acid]
= 7.14 + log [0.047 / 0.059]
pH = 7.04
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