When an Acid Solution Meets More Water...

BoJone: Reading this article requires a basic understanding of ionization equilibrium.

For the past two weeks, we have been studying the concepts of ionization equilibrium in the high school textbook Chemistry Elective 4 (People’s Education Press edition). Although we are in the “top class,” our progress remains quite slow. Regarding ionization equilibrium, a classmate once asked me a question:

When water is continuously added to an acid solution, why does the pH tend toward 7? (At standard temperature and pressure)

Clearly, this question is easy to understand because the \text{H}^+ ions are diluted after adding water. However, I am more interested in a question derived from this:

When water is continuously added to a (strong) acid solution, in which direction does the equilibrium shift?

Some friends might immediately think: after adding water, the concentration of hydrogen ions decreases, so the equilibrium should shift in the direction that increases the hydrogen ion concentration (the forward direction). Readers thinking this way have overlooked another phenomenon: water itself contains hydrogen ions and hydroxide ions. The hydroxide ions from the added water will cause the concentration of hydroxide ions in the mixture to increase. From that perspective, should the equilibrium shift in the direction that decreases the hydroxide ion concentration (the reverse direction)? Between these two contradictory conclusions, which one is right and which one is wrong?

Friends who have studied ionization know that strong acids are completely ionized. Therefore, in their aqueous solutions, only one ionization equilibrium exists:

It should be noted that this equilibrium exists in any aqueous solution. When the temperature is constant, the product of the concentrations of hydrogen ions and hydroxide ions divided by the concentration of water is a constant (where concentration refers to molar concentration), i.e.: \frac{c(\text{H}^+) \cdot c(\text{OH}^-)}{c(\text{H}_2\text{O})} = K

Since c(\text{H}_2\text{O}) \approx 1, the approximation c(\text{H}^+) \cdot c(\text{OH}^-) = K_w is usually regarded as a constant, known as the ion-product constant of water. To discuss our second question with relative precision, we will not use the approximate result.

Let us list the concentrations of the original acid, the added water, and the mixed solution at the instant of mixing:

Both the acid and the water reached equilibrium before mixing, meaning: \frac{c_1 c_2}{c_3} = \frac{c_4^2}{c_5} = K

To determine which way the equilibrium shifts after mixing, we must compare the value of \frac{c(\text{H}^+) \cdot c(\text{OH}^-)}{c(\text{H}_2\text{O})} with K. We can calculate: \begin{aligned} & \frac{c_1 V_1 + c_4 V_2}{V_1 + V_2} \cdot \frac{c_2 V_1 + c_4 V_2}{V_1 + V_2} \div \frac{c_3 V_1 + c_5 V_2}{V_1 + V_2} \\ & = \frac{V_1^2 c_1 c_2 + V_2^2 c_4^2 + V_1 V_2 (c_4 c_2 + c_1 c_4)}{V_1^2 c_3 + V_2^2 c_5 + V_1 V_2 (c_3 + c_5)} \end{aligned} \tag{1}

Since the original solution is acidic, we must have c_1 > c_4 > c_2. Therefore, (c_4 - c_2)(c_1 - c_4) > 0, from which we can derive: c_4 c_2 + c_1 c_4 > c_4^2 + c_1 c_2

Substituting this inequality along with c_1 c_2 = K c_3 and c_4^2 = K c_5 into equation (1), we obtain: \frac{c_1 V_1 + c_4 V_2}{V_1 + V_2} \cdot \frac{c_2 V_1 + c_4 V_2}{V_1 + V_2} \div \frac{c_3 V_1 + c_5 V_2}{V_1 + V_2} > K

Therefore, the reaction must proceed in the reverse direction (the direction of water formation).