Question

# A 3.670-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of...

A 3.670-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 20.0 mL of this solution was titrated with 0.07662-M NaOH. The pH after the addition of 23.98 mL of base was 5.90, and the equivalence point was reached with the addition of 44.00 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 =

moles NaOH at the equivalence point = 0.044 L x 0.07662 M =
= 0.00337

moles unknown acid in 20.0 mL= 0.00337

moles unknown acid in 100 mL = 0.00337 x 100 / 60=0.00561 = 5.61 mmoles

molar mass unknown acid = 3.670 g/0.00561 mol= 655.35g/mol

moles NaOH ( after the addition of 23.98 mL )= 0.02398 L x 0.07057 M=0.00183

moles acid = 3.670 g/ 655.35 g/mol=0.0056( in 100 mL)

moles acid in 60 mL = 0.0056 x 60/100=0.00336

HA + OH- => A- + H2O
moles acid = 0.00336 - 0.00183=0.00153
moles A- = 0.00183

total volume = 60 + 23.98 = 83.98 mL = 0.083 L

[HA]= 0.00153/ 0.083=0.0184 M
[A-]= 0.00183 / 0.083 =0.022 M

5.90 = pKa + log 0.022/ 0.0184 = pKa + 0.0784

pKa =5.82

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