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A 1.731-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of...

A 1.731-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 45.0 mL of this solution was titrated with 0.09322-M NaOH. The pH after the addition of 11.78 mL of base was 4.23, and the equivalence point was reached with the addition of 36.47 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 = _______

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Answer #1

A 1.731g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 45.0 mL of this solution was titrated with 0.09322M NaOH. The pH after the addition of 11.78 mL of base was 4.23, and the equivalence point was reached with the addition of 36.47 mL of base.

a) How many millimoles of acid are in the original solid sample?

b) What is the molar mass of the acid?

c) What is the pKa of the acid?

HA + NaOH NaA + H2O

3.4 1.1 0   

2.3 0 1.1

0 0
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