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Chapter 17: Problem 91

A sample of \(500 \mathrm{mg}\) of an unknown monoprotic acid was dissolved in \(50.0 \mathrm{~mL}\) of water and titrated with \(0.200 \mathrm{M}\) KOH. The acid required \(20.60 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After \(10.30 \mathrm{~mL}\) of base had been added in the titration, the pH was found to be 4.20 . What is the \(\mathrm{p} K_{a}\) for the unknown acid?

Short Answer

Expert verified
The molar mass of the unknown monoprotic acid is \(121.36 \mathrm{g/mol}\) and its \(\mathrm{p} K_{a}\) is 4.20.

Step by step solution

01

Calculate the moles of KOH required for titration to reach the equivalence point

The problem gives that 20.60 mL of 0.200 M KOH are required to reach the equivalence point. Using this information, we can now calculate the moles of KOH used in the titration: Moles of KOH = Volume of KOH (L) × Molarity of KOH = (20.60 mL × (1L / 1000 mL)) × 0.200 M = 0.00412 moles
02

Determine moles of the unknown acid

Since it is a monoprotic acid, at the equivalence point, the moles of the acid will be equal to the moles of the base. Therefore, the moles of the unknown acid are also 0.00412 moles.
03

Calculate the molar mass of the unknown acid

To find the molar mass of the unknown acid, we will use the given mass (500 mg) and the moles of the unknown acid calculated in step 2. Convert the mass to grams first. Mass of the unknown acid (g) = 500 mg × (1 g/1000 mg) = 0.5 g Molar mass of the unknown acid = mass of the unknown acid (g) / moles of the unknown acid Molar mass = 0.5 g / 0.00412 moles = 121.36 g/mol This is the molar mass of the unknown acid, which is the answer to part (a) of the problem.
04

Calculate the initial concentration of the unknown acid

We were given the initial volume of the solution as 50.0 mL. To find the initial concentration, we can use the moles of the unknown acid and the given volume. Initial concentration = moles of unknown acid / initial volume (L) Initial concentration = 0.00412 moles / (50.0 mL × (1 L / 1000 mL)) = 0.0824 M
05

Calculate the concentration of the unknown acid after adding 10.30 mL of 0.200 M KOH

Since 10.30 mL of 0.200 M KOH is exactly half of the volume required to reach the equivalence point, we are at the half-equivalence point of the titration. At this point, half of the unknown acid has reacted with KOH, and the remaining half is still in the initial form. Thus, the concentration of the unknown acid and its conjugate base is equal: Remaining moles of the unknown acid = 0.5 × 0.00412 moles = 0.00206 moles Concentration of the unknown acid and conjugate base = (0.00206 moles)/(50 mL + 10.30 mL) × (1 L / 1000 mL) = 0.0295 M
06

Calculate the pKa of the unknown acid using the pH after adding 10.30 mL of 0.200 M KOH

We were given that the pH at this point is 4.20. Since we are at the half-equivalence point, the pH is equal to the pKa: pH = pKa Thus, the pKa of the unknown acid is 4.20, which is the answer to part (b) of the problem.

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