Question

A sample of \(0.2140 \mathrm{~g}\) of an unknown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\) of water and titrated with \(0.0950 \mathrm{M}\) \(\mathrm{NaOH}\). The acid required \(30.0 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After \(15.0 \mathrm{~mL}\) of base had been added in the titration, the \(\mathrm{pH}\) was found to be \(6.50 .\) What is the \(K_{a}\) for the unknown acid?

Short answer

(a) The molar mass of the unknown monoprotic acid is 75.09 g/mol. (b) The Ka value of the unknown acid is 3.16 × 10⁻⁷.

Step-by-step solution

(a) Calculate the moles of NaOH used for titration

To find the moles of NaOH, we will use the concentration and volume of NaOH solution: Moles of NaOH = Concentration of NaOH × Volume of NaOH Moles of NaOH = \(0.0950 \mathrm{M} \times 30.0 \mathrm{~mL}\) Since the volume is in mL, we need to convert it to L by dividing by 1000: Moles of NaOH = \(0.0950 \mathrm{M} \times \frac{30.0 \mathrm{~mL}}{1000}\) Moles of NaOH = 0.00285 moles

Calculate the moles of the unknown acid

At the equivalence point, the number of moles of the acid (HA) is equal to the number of moles of the base (NaOH) since they react in a 1:1 ratio. Moles of acid = Moles of NaOH = 0.00285 moles

Calculate the molar mass of the unknown acid

To find the molar mass of the acid, we can use the given mass of the sample and the number of moles calculated earlier: Molar mass = \(\frac{Mass}{Moles}\) Molar mass = \(\frac{0.2140 \mathrm{~g}}{0.00285 \mathrm{~mol}}\) Molar mass = 75.09 g/mol Thus, the molar mass of the unknown monoprotic acid is 75.09 g/mol.

(b) Calculate the concentration of acid and base at the half-equivalence point

At the half-equivalence point, only half of the NaOH has been added. Thus, the volume of NaOH added is 15 mL, and the moles of NaOH added is: Moles of NaOH = Concentration of NaOH × Volume of NaOH Moles of NaOH = \(0.0950 \mathrm{M} \times \frac{15.0 \mathrm{~mL}}{1000}\) Moles of NaOH = 0.001425 moles Since we are given the pH of the solution at this point (pH = 6.50), we can find the concentration of H₃O⁺ ions: [H₃O⁺] = 10^(-pH) = 10^(-6.50) = 3.16 × 10⁻⁷ M Now, we can determine the concentration of the conjugate base, A⁻, which will be equal the moles of NaOH added: [A⁻] = 0.001425 moles

Calculate the concentration of the undissociated acid (HA) at the half-equivalence point

At the half-equivalence point, half of the acid has reacted with the base. The remaining concentration of the undissociated acid is: [HA] = moles of acid - moles of base = 0.00285 - 0.001425 = 0.001425 moles Since the total volume of the solution (acid + NaOH) at the half-equivalence point is 25.0 mL + 15.0 mL = 40.0 mL, we can convert the moles of HA to concentration: [HA] = \(\frac{0.001425 \mathrm{~mol}}{0.040 \mathrm{~L}}\) = 0.0356 M

Calculate the Ka value for the unknown acid

Using the concentrations of H₃O⁺, A⁻, and HA, we can calculate the Ka value of the acid: Ka = \(\frac{[H_{3}O^{+}][A^{-}]}{[HA]}\) Ka = \(\frac{(3.16 \times 10^{-7} \mathrm{~M})(0.0356 \mathrm{~M})}{0.0356 \mathrm{~M}}\) Ka = 3.16 × 10⁻⁷ The Ka value of the unknown acid is 3.16 × 10⁻⁷.

Key concepts

Equivalence Point

In an acid-base titration, the equivalence point is the stage at which the amount of titrant added is chemically equivalent to the substance in the solution being titrated. In simple terms, this is the point where the number of moles of acid equals the number of moles of base. This is crucial because it tells us that the acid has been completely neutralized by the base.

For example, in our given problem, the equivalence point is reached when 30.0 mL of NaOH is used. This is because NaOH is added to neutralize the unknown monoprotic acid, meaning that both reactants have reacted in a 1:1 ratio. Identifying this point accurately allows for the correct calculation of the molar mass, which depends on knowing exactly how many moles were involved in the reaction.

Equivalence points are often determined by noting a change in pH, which is commonly monitored using an indicator dye or a pH meter. It is not necessarily the same as the endpoint, which is the point where a visible change confirms the completion of the reaction, though they are ideally as close as possible.

Molar Mass Calculation

Calculating the molar mass of an unknown substance in a titration involves knowing two things: the mass of the sample and the moles of the substance involved in the reaction. The formula for molar mass is:\[\text{Molar Mass} = \frac{\text{Mass of Sample}}{\text{Moles of Substance}}.\]

In our example, we first need to establish the moles of NaOH added at the equivalence point, which matched the moles of the unknown acid due to their 1:1 reaction ratio. We found the moles of NaOH by multiplying its concentration (0.0950 M) by its volume (0.030 L), resulting in 0.00285 moles.

With a mass of 0.2140 g for the acid, we calculated its molar mass by dividing the mass by the number of moles (0.2140 g / 0.00285 mol), resulting in a molar mass of 75.09 g/mol. This calculation is crucial for determining the identity of the unknown acid by comparing the calculated molar mass with those of known acids.

Dissociation Constant (Ka)

The dissociation constant, represented as Ka, is a crucial value that shows the strength of an acid in a solution. It measures how well an acid dissociates into its ions at a given concentration. The larger the Ka, the stronger the acid, as it means more acid molecules are ionized.

For the unknown acid, calculating the Ka involved understanding the concentrations of hydrogen ions, the conjugate base, and the undissociated acid at the half-equivalence point. At this point, the concentration of the acid equals the base that has reacted with the titrant, simplifying the calculation drastically.

Given that the pH at the half-equivalence point was 6.50, we could calculate \[ [H_3O^+] = 10^{-pH} = 3.16 \times 10^{-7} \text{ M}. \]
The concentrations for both the conjugate base \( A^- \) and undissociated acid \( HA \) are crucial for calculating the Ka value using:

\[ \text{Ka} = \frac{[H_3O^+][A^-]}{[HA]}. \]

Substituting the values, we found Ka to be \( 3.16 \times 10^{-7} \), demonstrating the relatively weak acidic strength of the unknown monoprotic acid.