Question

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.

Short answer

In this problem, we assume the strong acid (HCl) reacts completely with the weak base (A⁻) before the weak acid (HA) dissociates. This simplifies the pH calculation by reducing the system to a weak acid solution. The final pH would not be significantly affected if the reaction order were different, as strong acids like HCl will always dominate the pH of the solution when mixed with weak acids or bases.

Step-by-step solution

Identify which ions are present in the final solution and their initial concentrations

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)

Assume the strong acid (HCl) reacts first with the weak base (A⁻)

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.

Calculate the remaining concentrations of HA and A⁻ from the weak acid

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.

Calculate the pH using the Henderson-Hasselbalch equation

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⁺]).

Calculate the pH using the concentration of 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.

Conclusion

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.

Key concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a valuable tool in chemistry for calculating the pH of buffer solutions. It provides a simple formula:

• \[\text{pH} = \text{p}K_\text{a} + \log \frac{[\text{A}^{-}]}{[\text{HA}]}\]

This equation relates the pH to the strength of the acid (given by its dissociation constant, \(K_a\)) and the ratio of the concentrations of the base form ([A⁻]) and the acid form ([HA]).
When both forms are equally concentrated, the pH equals the pKₐ of the acid. However, as seen in the exercise, if the concentration of A⁻ becomes zero, other methods must be used to calculate pH as dividing by zero isn't possible.

Weak Acid

A weak acid partially dissociates in water, unlike strong acids that dissociate completely. In our example, HA is the weak acid with a given dissociation constant, \(K_a = 1.0 \times 10^{-6}\).
This relatively small value of \(K_a\) indicates that the acid ionizes only slightly in solution.
Understanding the behavior of weak acids is crucial for buffer solutions which help maintain a stable pH even when small amounts of other acids or bases are added.

• They form equilibrium with their conjugate base, represented as HA ⇌ H⁺ + A⁻.
• The extent of dissociation and the concentration of hydrogen ions (H⁺) are important for pH calculations.

Strong Acid

Contrary to weak acids, a strong acid like HCl dissociates completely in solution.
This means that in a solution of HCl, nearly all the acid molecules break down to release H⁺ ions.
These ions react rapidly with any available base ions, such as A⁻, from our buffer solution.

• Complete dissociation simplifies calculations as all the acid is assumed to convert into ions.
• In buffer problems, the complete dissociation is used to adjust concentrations before further calculations with weak acids.

Understanding strong acids' behavior helps us predict how they will alter the pH of weak acid solutions, often significantly lowering it.

pH Calculation

The pH calculation in this exercise involves several steps and assumptions.
Initially, we observe the complete reaction of the strong acid with the base form of the buffer, leading to no base remaining for further reactions.
With no A⁻ after the strong acid's reaction, we then rely on the dissociation of the weak acid to determine the pH.

• We use the expression for \(K_a\): \[K_\text{a} = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}, \] replacing A⁻ with a variable \(x\), representing [H⁺].
• Solving for \(x\) allows determination of [H⁺], which is then converted to pH using \[ \text{pH} = -\log [\text{H}^+]\. \]

It is crucial to approach these calculations systematically, ensuring each step is carried out with precision to achieve accurate results.