A 0.5224-g sample of an unknown monoprotic acid was titrated with 0.0998 M NaOH.

Question

A 0.5224-g sample of an unknown monoprotic acid was titrated with 0.0998 M NaOH.

The equivalence point of the titration occurred at 23.82 mL.

Determine the molar mass of the unknown acid.

Please explain the Equivalence point too.

Answer 1

Since I had difficulty reading the previous answer, I thought I'd try a different approach.

First, the equivalence point is when the moles of acid (HA) is equal to the moles of base. At that point all of the monoprotic acid (HA) is used up, and all of the base is used up, and you end up with water and the conjugate base (A^-) of the monoprotic acid. That's the only explanation I can provide.

In the current problem

moles of base (NaOH) = 23.82 ml x 1 L / 1000 ml x 0.0998 mols / L = 0.002377 moles NaOH

moles of acid (HA) = 0.002377 moles since this is at the equivalence point.

The molar mass of the acid is defined as grams of acid / mole of acid

molar mass = 0.5224 g / 0.002377 moles

molar mass = 219.8 g / mole

Answer 2

Step 1: Write the reaction for the titration

Since the unknown acid is monoprotic (releases one proton per molecule), it reacts with NaOH in a 1:1 molar ratio:

HA(aq)+OH−(aq)→H2O(l)+A−(aq)HA (aq) + OH^- (aq) \rightarrow H_2O (l) + A^- (aq)HA(aq)+OH−(aq)→H2​O(l)+A−(aq)Step 2: Calculate the moles of NaOH used

We know the volume and molarity of the NaOH solution at the equivalence point, so we can calculate the moles of NaOH used:

moles of NaOH=Molarity of NaOH×Volume of NaOH (in L)\text{moles of NaOH} = \text{Molarity of NaOH} \times \text{Volume of NaOH (in L)}moles of NaOH=Molarity of NaOH×Volume of NaOH (in L) moles of NaOH=0.0998 M×23.82 mL×11000 L/mL\text{moles of NaOH} = 0.0998 \, \text{M} \times 23.82 \, \text{mL} \times \frac{1}{1000} \, \text{L/mL}moles of NaOH=0.0998M×23.82mL×10001​L/mL moles of NaOH=0.0998×0.02382=0.00238 moles\text{moles of NaOH} = 0.0998 \times 0.02382 = 0.00238 \, \text{moles}moles of NaOH=0.0998×0.02382=0.00238molesStep 3: Moles of the unknown acid

Since the reaction is 1:1, the moles of the unknown acid (HA) will be equal to the moles of NaOH at the equivalence point:

moles of HA=0.00238 moles\text{moles of HA} = 0.00238 \, \text{moles}moles of HA=0.00238molesStep 4: Calculate the molar mass of the unknown acid

The molar mass of the unknown acid is given by the formula:

Molar mass of acid=Mass of acidMoles of acid\text{Molar mass of acid} = \frac{\text{Mass of acid}}{\text{Moles of acid}}Molar mass of acid=Moles of acidMass of acid​We are given the mass of the acid sample as 0.5224 g, so:

Molar mass of acid=0.5224 g0.00238 moles\text{Molar mass of acid} = \frac{0.5224 \, \text{g}}{0.00238 \, \text{moles}}Molar mass of acid=0.00238moles0.5224g​ Molar mass of acid=219.3 g/mol\text{Molar mass of acid} = 219.3 \, \text{g/mol}Molar mass of acid=219.3g/molFinal Answer:

The molar mass of the unknown monoprotic acid is 219.3 g/mol.