Consider a solution formed by mixing 100.0 \(\mathrm{mL}\) of 0.10 \(\mathrm{M}\) $\mathrm{HA}\left(K_{\mathrm{a}}=1.0 \times 10^{-6}\right), 100.00 \mathrm{mL}\( of \)0.10 \mathrm{M} \mathrm{NaA},\( and 100.0 \)\mathrm{mL}$ of 0.10 \(\mathrm{M} \mathrm{HCl} .\) In calculating the pH for the final solution, you would make some assumptions about the order in which various reactions occur to simplify the calculations. State these assumptions. Does it matter whether the reactions actually occur in the assumed order? Explain.

In the final solution, the following ions will be present: - H⁺ from HA (100 mL × 0.10 M) - A⁻ from HA (100 mL × 0.10 M) - A⁻ from NaA (100 mL × 0.10 M) - Na⁺ from NaA (100 mL × 0.10 M) - H⁺ from HCl (100 mL × 0.10 M) - Cl⁻ from HCl (100 mL × 0.10 M)

We will assume that the strong acid, HCl, reacts completely with the weak base, A⁻, forming HA and Cl⁻. Since both HCl and A⁻ have the same initial concentration (0.10 M), the reaction will consume all HCl and A⁻ from NaA: H⁺ (from HCl) + A⁻ (from NaA) → HA After this reaction, all the HCl will have reacted, and the concentration of A⁻ from NaA will have reduced to zero.

Since the initial concentration of HA and A⁻ are both 0.10 M, and the total volume of the solution is now 300 mL, the final concentrations will be: \[C_{HA} = \frac{100\, \mathrm{mL} \times 0.10\, \mathrm{M} + 100\, \mathrm{mL} \times 0.10\, \mathrm{M}}{300\,\mathrm{mL}} = 0.10\, \mathrm{M}\] \[C_{A^-} = \frac{100\, \mathrm{mL} \times 0.10\, \mathrm{M} - 100\, \mathrm{mL} \times 0.10\, \mathrm{M}}{300\,\mathrm{mL}} = 0\,\mathrm{M}\] Now, we have a solution of a weak acid (HA) at a concentration of 0.10 M.

As we now have a solution of a weak acid, we can calculate the pH using the Henderson-Hasselbalch equation: \[\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\] However, since the concentration of A⁻ is zero and we cannot divide by zero, we need to use the following general equation to calculate the [H⁺] concentration in a weak acid solution: \[K_\mathrm{a} = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]}\] Since [A⁻] = 0, we can assume that a small amount of HA will dissociate into H⁺ and A⁻. Let x be the concentration of H⁺ that dissociates from HA. The equation becomes: \[K_\mathrm{a} = \frac{x^2}{[HA]-x}\] Now substitute the given value of Ka (1.0 × 10⁻⁶) and the calculated concentration of HA (0.10 M) to find x ([H⁺]).

Finally, calculate the pH using the formula: \[\mathrm{pH} = -\log [\mathrm{H}^+]\] Plug the computed [H⁺] value from the previous step into this formula to find the final pH of the solution.

Making assumptions about the order of reactions helped to simplify the calculation of the pH in this exercise. Furthermore, since the strong acid (HCl) reacts with the weak base (A⁻) completely before the weak acid (HA) dissociates, the final pH would not be significantly affected if the reaction order were different.