Question

Four grams of a monoprotic weak acid are dissolved in water to make \(250.0 \mathrm{~mL}\) of solution with a \(\mathrm{pH}\) of \(2.56 .\) The solution is divided into two equal parts, \(A\) and \(B\). Solution \(A\) is titrated with strong base to its equivalence point. Solution \(\mathrm{B}\) is added to solution A after solution \(\mathrm{A}\) is neutralized. The \(\mathrm{pH}\) of the resulting solution is \(4.26 .\) What is the molar mass of the acid?

Short answer

The molar mass of the weak acid is approximately \(116.3 \, \mathrm{g/mol}\).

Step-by-step solution

Calculate the concentration of hydronium ions

Use the given pH to find the concentration of hydronium ions using the formula: \([\mathrm{H_3O^+}] = 10^{-\mathrm{pH}}\) \([\mathrm{H_3O^+}] = 10^{-2.56} \approx 2.74 \times 10^{-3} \, \mathrm{M}\)

Calculate the dissociation constant (Ka) for the weak acid

Use the given neutralization of solution A at the equivalence point to calculate Ka: \(\mathrm{K_a} = \frac{[\mathrm{H_3O^+}][\mathrm{A^{-}}]}{[\mathrm{HA}]} = \frac{[\mathrm{H_3O^+}]^2}{[\mathrm{HA}] - [\mathrm{H_3O^+}]}\) At the equivalence point, \([\mathrm{HA}] - [\mathrm{H_3O^+}] = [\mathrm{A^{-}}]\): \(\mathrm{K_a} = [\mathrm{H_3O^+}] = 2.74 \times 10^{-3}\)

Determine the moles of the weak acid

Use the total mass of the weak acid (4g) and its molarity in the initial solution (0.5 times the initial concentration) to determine the moles of the weak acid in the initial solution: Moles of weak acid = \(\frac{4\,\mathrm{g}}{\mathrm{Molar\,Mass}} = \frac{1}{2} \cdot 0.25\,\mathrm{L} \cdot [\mathrm{HA_0}]\)

Calculate the concentration of the weak acid in Solution B

Use the moles of weak acid to find the concentration of the weak acid in Solution B: \([\mathrm{HA_B}] = \frac{\text{moles of weak acid}}{0.125 \, \mathrm{L}} = 2\,[\mathrm{HA_0}]\)

Calculate the concentrations of hydronium ions and weak acid in the final solution

Use the pH of the final solution to find the concentration of hydronium ions and weak acid (HA) in the final solution, after Solution B is mixed with neutralized Solution A: \([\mathrm{H_3O^+}]_\mathrm{final} = 10^{-4.26} \approx 5.49 \times 10^{-5} \, \mathrm{M}\) Since \(\mathrm{HA_B}\) is mixed with the same volume of neutralized solution A, \([\mathrm{HA}]_\mathrm{final} = [\mathrm{HA}]/2 = [\mathrm{HA_B}]\)

Calculate the molar mass of the weak acid

Use the concentration of hydronium ions and weak acid in the final solution, along with Ka, to solve for the molar mass of the weak acid: \(\mathrm{K_a} = \frac{[\mathrm{H_3O^+}]_\mathrm{final}\,[\mathrm{A^{-}}]_\mathrm{final}}{[\mathrm{HA}]_\mathrm{final}} \Rightarrow [\mathrm{HA}]_\mathrm{final} = [\mathrm{HA_B}] = \frac{[\mathrm{H_3O^+}]_\mathrm{final}^2}{\mathrm{K_a}}\) Plugging the numbers from Steps 1-5, we get: \(2 \, [\mathrm{HA_0}] = \frac{(5.49 \times 10^{-5})^2}{2.74 \times 10^{-3}}\) Solving for \([\mathrm{HA_0}]\): \([\mathrm{HA_0}] \approx 5.49 \times 10^{-5} \, \mathrm{M}\) Now, use the previous relationship between the mass and moles of the weak acid (from Step 3) to find the molar mass: Molar Mass = \(\frac{4\,\mathrm{g}}{0.5 \cdot 0.125\,\mathrm{L} \cdot [\mathrm{HA_0}]}\) Molar Mass \(\approx 116.3\, \mathrm{g/mol}\) The molar mass of the weak acid is approximately \(116.3 \, \mathrm{g/mol}\).