A 6.053-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of s

Question

A 6.053-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 15.0 mL of this solution was titrated with 0.08386-M NaOH. The pH after the addition of 23.33 mL of base was 3.90, and the equivalence point was reached with the addition of 43.69 mL of base.

a) How many millimoles of acid are in the original solid sample? Hint: Don't forget the dilution. Correct: Your answer is correct. mmol acid

b) What is the molar mass of the acid? Incorrect: Your answer is incorrect. g/mol

c) What is the pKa of the acid? pKa =

Explanation / Answer

a)

mol of NaOH = MV = 43.69 * 0.08386 = 3.6638 mmol of base

if this is monoprotic, then

mmol of acid = 3.6638 mmol

this is present in 15 mL

total 100 mL --> 100/15*3.6638 = 24.425 mmol of acid in 100 mL

b)

Molar Mass = mas s/ mol = (6.053)/(24.425*10^-3) = 247.81 g/mol

c)

pKa an be obtained via

pH = pKa + log(A-/HA)

so,

initially mmol of HA = 3.6638

after adding --> 23.33 ml of base

mmol of bas eadded = MV = 23.33*0.08386 = 1.9564

mmol of HA left =3.6638 -1.9564 = 1.7078

mmol of A- formed = 0+ 1.9564

then

we know pH =3.90

pH = pKa + log(A-/HA)

3.90 = pKa + log(1.9564/1.7078)

pKa = 3.90 -  log(1.9564/1.7078)

pKa = 3.8409

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