A 5.344-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of s

Question

A 5.344-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 0.08158-M NaOH. The pH after the addition of 25.13 mL of base was 3.77, and the equivalence point was reached with the addition of 39.64 mL of base.

a) How many millimoles of acid are in the original solid sample? Hint: Don't forget the dilution. Incorrect: Your answer is incorrect. 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 equivalence point moles of monoprotic acid = base NaOH moles

NaOH moles = M x V ( in liters) = 0.08158 x 39.64 /1000 = 0.00323383

Acid moles present in 15 ml of solution = 0.00323383

Acid Conc = Moles/vol in L = 0.00323383 /( 0.015) = 0.2156

Now initially we had 100 ml solution

So moles of acid for 100 ml = M x V ( in liters) = 0.2156 x 0.1 = 0.02156

Milli miles = Moles x 1000 = 0.02156 x 1000 = 21.56 nearly

b) Moles = Mass in grams / Molar mass

0.02156 = 5.344 / Molar mass

Molar mass = 247.87 g/mol

c) when 25.13 ml base used i.e OH- moles = M x V = 0.08518 x 25.13 /1000 = 0.00205

acid moles after reaction with OH- = 0.02156-0.00205 = 0.01951

conjugate base moles = base moles = 0.00205

Now total vol = 15+25.13 = 40.13 ml = 0.04013 L

No pH = pka + log [Conjugate base] /[acid]

3.77 = pka + log ( 0.00205/0.04013) / ( 0.01951/0.04013)

3.77 = pka + log ( 0.00205/0.01951)

pka = 4.75

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