a 25.00 ml sample of a weak acid is titrated with 0.225 M NaOH. a total...
Homework Answers
assume weak acid = HA
millimoles of NaOH added = 12.20 x 0.225 = 2.745
2.745 millimoles acid reacted
salt formed = 2.745
total volume = 25.00 + 12.20 = 37.2 mL
[salt] = 2.745 / 27.2 = 0.10 M
for salt of weak acid
pH = 1/2 [pKw + pKa + log C]
pH = 9.96
C = 0.10 M
9.96 = 1/2 [14 + pKa + log 0.10]
19.92 = 14 + pKa - 1
19.92 = 13 + pKa
pKa = 6.92
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