a) initially only benzoic acid present
for pH of weak acids we have direct formula
pH = 1/2 [pKa - log C]
pKa = - log Ka = - log 6.5 x 10-5 = 4.19
C = 0.150 M
pH = 1/2 [4.19 - log 0.150]
pH = 2.51
b) half equivalence point means 50% acid convert to salt
at half equivalence point we have pH = pKa
so
pH = 4.19
c) millimoles of acid = 40.0 x 0.150 = 6.0
6.0 millimoles KOH must be added to reach equivalence point
6.0 = V x 0.300
V = 20.0 mL
total volume = 40 + 20 = 60 mL
[salt] = 6.0 / 60 = 0.10 M
at equivalence point
pH = 1/2 [pKw + pKa + log C]
pH = 1/2 [14 + 4.19 + log 0.10]
pH = 8.60
Quest. 7 (30 pts). Benzoic acid (C6H5COOH), 40.0 mL of 0.150 M, is being titrated with 0.300 M potassium hydroxide (KOH...
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