A sample of a certain monoprotic weak acid was dissolved in water and titrated with 0.125\(M \mathrm{NaOH}\) , requiring 16.00 \(\mathrm{mL}\) to reach the equivalence point. During the titration, the pH after adding 2.00 \(\mathrm{mL}\) . NaOH was 6.912 . Calculate \(K_{\mathrm{a}}\) for the weak acid.

To determine the number of moles of NaOH that reacted, we can use the volume and molarity given: Moles of NaOH = Molarity × Volume Moles of NaOH = 0.125 M × 16.00 mL × \(\frac{1\,\mathrm{L}}{1000\,\mathrm{mL}}\) Moles of NaOH = 0.00200 moles Since the weak acid is monoprotic, the moles of weak acid will equal the moles of NaOH at the equivalence point. Moles of weak acid = Moles of NaOH = 0.00200 moles

The reaction between the weak acid (HA) and NaOH can be represented by the following balanced equation: HA + OH⁻ → A⁻ + H₂O

The number of moles of NaOH added when 2.00 mL is used can be calculated as follows: Moles of NaOH added at 2.00 mL = Molarity × Volume Moles of NaOH added at 2.00 mL = 0.125 M × 2.00 mL × \(\frac{1\,\mathrm{L}}{1000\,\mathrm{mL}}\) Moles of NaOH added at 2.00 mL = 0.00025 moles Using the balanced chemical equation, the moles of weak acid that reacted are equal to the moles of NaOH added at 2.00 mL: Moles of HA reacted = 0.00025 moles The remaining moles of the weak acid (HA), and the moles of the conjugate base (A⁻) formed can be calculated as: Moles of HA remaining = Initial moles of HA - Moles of HA reacted Moles of HA remaining = 0.00200 moles - 0.00025 moles = 0.00175 moles Moles of A⁻ formed = Moles of HA reacted = 0.00025 moles Now we will determine the total volume of the solution at this point: Total volume = Initial volume of weak acid + Volume of NaOH added Total volume = 16.00 mL + 2.00 mL = 18.00 mL To calculate the concentration of HA and A⁻ at this point, we will divide the moles by the total volume: Concentration of HA = \(\frac{0.00175\,\mathrm{moles}}{0.018\,\mathrm{L}}\) = 0.09722 M Concentration of A⁻ = \(\frac{0.00025\,\mathrm{moles}}{0.018\,\mathrm{L}}\) = 0.01389 M

We will create an ICE table for the weak acid ionization: | | HA | H₂O (l) | H₃O⁺ | A⁻ | |------- |------------|----------|-----------|-----------| | I | 0.09722 M | -- | 0 M | 0.01389 M | | C | -x | -- | +x | +x | | E | 0.09722-x M | -- | x M | 0.01389+x M |

We know the pH after adding 2.00 mL of NaOH is 6.912, so we can find the concentration of H₃O⁺ ions at this point: pH = -log[H₃O⁺] 6.912 = -log[H₃O⁺] Solve for [H₃O⁺]: [H₃O⁺] = 10^(-6.912) = 1.2 × 10^(-7) M Now, we can use the equilibrium concentrations from the ICE table to find Ka for the weak acid: Default variables: x = [H₃O⁺] = 1.2 × 10^(-7) M 0.09722 - x ≈ 0.09722 M (since x is very small) Using the ionization equation for the weak acid: Ka = \(\frac{[A⁻][H₃O⁺]}{[HA]}\) Ka = \(\frac{(0.01389 + x)(1.2 × 10^{-7})}{0.09722}\) Solve for Ka: Ka = 1.72 × 10^(-5) Thus, the acid dissociation constant for the weak acid is approximately 1.72 × 10^(-5).