A student titrates an unknown weak acid, HA, to a pale pink phenolphthalein end point with \(25.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\). The student then adds \(13.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}\). The \(\mathrm{pH}\) of the resulting solution is \(4.70\). How is the value of \(\mathrm{p} K_{\mathrm{a}}\) for the unknown acid related to 4.70?

In this problem, there are three important stages to consider: 1. The titration of the weak acid with NaOH until the phenolphthalein endpoint. 2. The addition of HCl to the solution. 3. Calculating the pH of the resulting solution and its relationship with the pKa of the unknown weak acid.

First, we need to calculate the moles of the weak acid in the solution. We are given the volume (25.0 mL) and concentration (0.100 M) of NaOH used to titrate the weak acid to the endpoint, where all the weak acid is neutralized. To find the moles of weak acid, we can use the following equation: moles of weak acid = moles of NaOH = volume_NaOH × concentration_NaOH moles of weak acid = \(25.0 \times 10^{-3} \mathrm{L} \times 0.100 \mathrm{M}\) = \(2.50 \times 10^{-3} \mathrm{mol}\) At the endpoint of the titration, the weak acid, HA, has been completely neutralized, and the only species present is its conjugate base, A⁻.

Now we will consider the addition of 13.0 mL of 0.100 M HCl to the solution. The moles of HCl can be calculated in a similar way to the moles of the weak acid: moles of HCl = volume_HCl × concentration_HCl moles of HCl = \(13.0 \times 10^{-3} \mathrm{L} \times 0.100 \mathrm{M}\) = \(1.30 \times 10^{-3} \mathrm{mol}\) When the HCl is added, it reacts with the conjugate base, A⁻, to form the weak acid, HA. Since we have more moles of A⁻ than HCl, not all A⁻ will be consumed: moles of A⁻ remaining = moles of A⁻ initially - moles of HCl = \(2.50 \times 10^{-3} - 1.30 \times 10^{-3} = 1.20 \times 10^{-3} \mathrm{mol}\) moles of HA formed = moles of HCl = \(1.30 \times 10^{-3} \mathrm{mol}\)

After the addition of HCl, the solution contains both the weak acid, HA, and its conjugate base, A⁻. Since we know the moles of each, we can calculate their concentrations by dividing by the total volume of the solution, which is 25.0 mL + 13.0 mL = 38.0 mL: \[ [\mathrm{A^-}] = \frac{1.20 \times 10^{-3} \mathrm{mol}}{38.0 \times 10^{-3} \mathrm{L}} = 0.0316 \mathrm{M} \] \[ [\mathrm{HA}] = \frac{1.30 \times 10^{-3} \mathrm{mol}}{38.0 \times 10^{-3} \mathrm{L}} = 0.0342 \mathrm{M} \] Now, we can write an expression for the Ka, the acid dissociation constant, for the weak acid: \[ K_{\mathrm{a}} = \frac{[\mathrm{H^+}] [\mathrm{A^-}]}{[\mathrm{HA}]} \] Since the pH of the solution is given as 4.70, we can find the concentration of H⁺ ions: \[ [\mathrm{H^+}] = 10^{-\mathrm{pH}} = 10^{-4.70} = 1.995 \times 10^{-5} \mathrm{M} \] Now, we can plug in the concentrations we found earlier into our Ka expression: \[ K_{\mathrm{a}} = \frac{(1.995 \times 10^{-5})(0.0316)}{0.0342} = 1.841 \times 10^{-6} \] Finally, we can find the relationship between the Ka and the given pH by calculating pKa: \[ \mathrm{p} K_{\mathrm{a}} = -\log_{10}(K_{\mathrm{a}}) = -\log_{10}(1.841 \times 10^{-6}) = 5.735 \] So, the pKa of the unknown weak acid, HA, is 5.735, which shows that it is less acidic than the final pH of the solution, which is 4.70.