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 \(4.7\). How is the value of \(\mathrm{p} K_{\mathrm{a}}\) for the unknown acid related to \(4.7 ?\)

First, we need to find the moles of NaOH and HCl. We can use the given volumes and molarity to do this. Moles of NaOH: \(moles = Molarity \times Volume\) \(moles_{NaOH} = 0.100 \mathrm{M} \times 25.0 \mathrm{mL} = 0.00250 \mathrm{mol}\) Moles of HCl: \(moles_{HCl} = 0.100 \mathrm{M} \times 13.0 \mathrm{mL} = 0.00130 \mathrm{mol}\) Since the weak acid (HA) is titrated with the NaOH completely, the amount of moles HA is equal to the amount of moles of NaOH. \(moles_{HA} = moles_{NaOH} = 0.00250 \mathrm{mol}\) We also have an excess of strong acid, HCl. The amount of remaining HCl can be found by subtracting the moles of NaOH from the moles of HCl: \(moles_{HCl_{excess}} = moles_{HCl} - moles_{NaOH} = 0.00130 \mathrm{mol} - 0.00250 \mathrm{mol} = -0.00120 \mathrm{mol}\) Since we cannot have negative moles, we must have made an error, and it must be that moles of NaOH = moles of HA + moles of HCl. So: \(moles_{HA} = moles_{NaOH} - moles_{HCl} = 0.00250 \mathrm{mol} - 0.00130 \mathrm{mol} = 0.00120 \mathrm{mol}\) Now we have the correct moles of HA and excess HCl.

The pH of the resulting solution is given as 4.7. We can use this value to find the concentration of H+ ions: \[H^+ = 10^{-pH} = 10^{-4.7} = 1.995\times10^{-5} \mathrm{M}\]

The dissociation of the weak acid HA into H+ and A- ions can be written as: \( HA \rightleftharpoons H^+ + A^-\) We can write the expression for the equilibrium constant, \(K_a\): \[K_a = \frac{[H^+][A^-]}{[HA]}\] Since we have the data for the concentration of HCl, we can calculate the resulting concentration of A- ions and HA: Concentration of HA: \([HA] = \frac{moles_{HA}}{Total \: volume} = \frac{0.00120 \: \mathrm{mol}}{25.0 \: \mathrm{mL} + 13.0 \: \mathrm{mL}} = 3.2258 \times 10^{-5} \: \mathrm{M}\) Concentration of A- ions: \([A^-] = \frac{moles_{HCl_{excess}}}{Total \: volume} = \frac{0.00130 \: \mathrm{mol}}{25.0 \: \mathrm{mL} + 13.0 \: \mathrm{mL}} = \frac{0.00130 \: \mathrm{mol}}{38.0 \: \mathrm{mL}} = 3.4211 \times 10^{-5} \: \mathrm{M}\) Now we can substitute these values into the \(K_a\) expression: \[K_a = \frac{(1.995\times10^{-5} \: \mathrm{M})(3.4211\times10^{-5} \: \mathrm{M})}{(3.2258\times10^{-5} \: \mathrm{M})} = 2.128 \times 10^{-5}\] Finally, we can find the \(\mathrm{p}K_a\): \[\mathrm{p} K_a = -\log_{10}(K_a) = -\log_{10}(2.128 \times 10^{-5}) = 4.672\] To summarize, the value of \(\mathrm{p}K_a\) for this unknown weak acid is related to the pH of the resulting solution through the expression \( \mathrm{p}K_a = 4.672 \), which implies that the weak acid has a lower dissociation constant than the final pH, making it a weaker acid than the final solution.