Question

A 1.33 g sample of a weak monoprotic acid, HA, was dissolved in distilled water. The HA solution was then titrated with 0.224 M NaOH. The equivalence point was reached when 45.0 mL of the NaOH solution had been added.

a) Identify the titrant.
b) Identify the analyte.
c) Write the balanced net-ionic equation for the reaction that occurs when the solution of NaOH is added to the solution of HA (a generic weak acid).
d. Calculate the number of moles of HA from the titration data.
e. Calculate the molar mass of HA.

Answer 1

The titrant in the scenario described is 0.224 M NaOH, and the analyte is the weak monoprotic acid HA. The balanced net-ionic equation is HA(aq) + OH⁻(aq) → A⁻(aq) + H₂O(l). Using the titration data, the molar mass of HA is calculated to be 131.94 g/mol.

Step-by-step explanation:

A) The titrant is the substance added from the burette to react with the analyte, which in this case is 0.224 M NaOH (sodium hydroxide).

B) The analyte is the substance being analyzed or tested, which is the weak monoprotic acid HA.

C) The balanced net-ionic equation for the reaction of NaOH (a strong base) with HA (a weak acid) is:
HA(aq) + OH⁻(aq) → A⁻(aq) + H₂O(l).

D) To calculate the number of moles of HA from the titration data, use the formula:

n = C x V

where C is the concentration of NaOH (0.224 M) and V is the volume of NaOH added at the equivalence point (45.0 mL or 0.045 L). The calculation is:

n = 0.224 M x 0.045 L = 0.01008 mol

E) To calculate the molar mass of HA, use the mass of the sample and the number of moles:

Molar mass = mass / moles

Molar mass = 1.33 g / 0.01008 mol = 131.94 g/mol