Question

# A 2.211-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of...

A 2.211-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 30.0 mL of this solution was titrated with 0.06886-M NaOH. The pH after the addition of 25.59 mL of base was 4.22, and the equivalence point was reached with the addition of 41.47 mL of base.

a) How many millimoles of acid are in the original solid sample? Hint: Don't forget the dilution.

mmol acid

b) What is the molar mass of the acid?

g/mol

c) What is the pKa of the acid?

pKa =

at the equivalence point millimoles of acid = millimoles of base

C x 30 .0 = 0.06886 x 41.47

C = 0.09519 M

molarity of of acid = 0.09519 M

a)

millimoles of acid = 0.09519 x 100

= 9.52

b)

molarity = moles x 1000 / 100

0.09519 M = moles x 10

moles = 9.52 x 10^-3

molar mass = mass / moles

= 2.211 / 9.52 x 10^-3

= 232.2 g/mol

c)

millimoles base = 0.06886 x 25.59 = 1.762

acid millimoles = 0.09519 x 30 = 2.856

acid   +   base --------------> salt

2.856 1.762    0

1.094 0 1.762

pH = pKa + log [salt /acid]

4.22 = pKa + log (1.762 / 1.094)

pKa = 4.01

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