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Question- Determine the volume of 0.150M sodium benzoate needed to have a pH 3.72 for 0.0500 mol of benzoic acid (C6H5COOH). pKa= 4.202
pH OF BUFFER SOLUTION
3.72 = pKa+ log[salt]/[acid]
3.72 = 4.202 + log(0.15/x)
X= concentration of Benzoicacid = 0.455 M <--- how did you get this number!?!?!
Molarity M = No of moles of solute/ volume of solution in liters
V= 0.05/0.4550 = 0.11 liter = 110 ml
Give reasons or calculations to sustain your answer.
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