Question

Calculate the \(\mathrm{pH}\) of a \(0.5 \mathrm{M}\) solution of \(\mathrm{Ca}(\mathrm{HSe})_{2}\), given that \(\mathrm{H}_{2}\) Se has \(K_{\mathrm{a}_{1}}=1.3 \times 10^{-4}\) and \(K_{\mathrm{a}_{2}}=1 \times 10^{-11}\)

Short answer

The pH of the 0.5M solution of Ca(HSe)2 is 2.24.

Step-by-step solution

Identify the Acid Ionization Constants

The problem supplies two acid ionization constants Ka1 and Ka2. Ka1 corresponds to the first ionization of the acid i.e. \( \mathrm{Ca}(\mathrm{HSe})_{2} \rightarrow \mathrm{HSe^{-}}+ \mathrm{CaHSe^{+}} \), which has a value of \(1.3 \times 10^{-4}\). Ka2 corresponds to the second ionization i.e. \( \mathrm{HSe^{-}} \rightarrow \mathrm{Se^{2-}}+ \mathrm{H^{+}} \), which has a value of \(1 \times 10^{-11}\)

Calculating the Concentration of H+ from first ionization

After the first ionization, the concentration of HSe- is equal to the initial concentration of Ca(HSe)2 which is 0.5 M. Because Ka1 is significantly larger than Ka2, we can assume that all ionization comes from the first ionization stage. Using the formula Ka1 = [H+][HSe-]/[Ca(HSe)2], and assuming [H+] equals [HSe-] because for each Ca(HSe)2 ionized one H+ is released, we can calculate [H+]. Solving the equation for [H+] gives [H+] = sqrt(Ka1 x [Ca(HSe)2]) = sqrt((1.3 x 10^-4) x (0.5M)) = 0.0057 M.

Calculating the pH

The pH of a solution is calculated using the formula pH = -log[H+]. Substituting the calculated [H+] from step 2 into this equation gives pH = -log(0.0057) = 2.24

Key concepts

Acid Ionization Constants

Understanding acid ionization constants is crucial for interpreting how acids behave in solution. These constants, denoted as Ka, provide insight into the strength of an acid by quantifying its tendency to donate protons (H+) to water, creating hydronium ions (H3O+). A higher Ka value indicates a stronger acid, as it means a greater proportion of the acid molecules dissociate to release H+ ions.

Take, for example, a diprotic acid like H2Se, which can donate two protons. It has two dissociation steps, each with its own Ka value. The first dissociation constant, Ka1, reflects the strength of the acid to donate the first proton and is generally higher than Ka2, the constant for the second dissociation. This is because, after the first proton is given off, the resulting anion is less likely to release its proton due to the negative charge repulsion.

When calculating H+ concentration, we often focus on the first ionization constant if it is significantly larger than the second, which allows for a simplified approach to understanding the acid behavior.

Acid-Base Equilibria

The acid-base equilibria are the balance points in which acids and bases in solution exist in both their protonated and deprotonated forms. It is essential to understand that most acid-base reactions are reversible, reaching a state where the forward and reverse reactions occur at the same rate, which we describe as an equilibrium.

In the context of H2Se dissociation, equilibria are established for each proton donation process. The equilibrium predominantly lies to the right for the first dissociation due to a higher Ka1, suggesting that H2Se effectively loses its first proton to form HSe-. However, for the second dissociation, with a much smaller Ka2, the equilibrium lies substantially to the left, indicating that HSe- does not readily lose its remaining proton.

This concept is fundamental when predicting the extent to which an acid will affect the pH of a solution, as it helps to determine the concentration of H+ ions in solution at equilibrium.

Concentration of H+

The concentration of H+ ions directly impacts the acidity of a solution. Calculating this concentration is essential for predicting the pH. For acids that are not 100% dissociated, like H2Se, the concentration of H+ is not simply the stoichiometric concentration of the acid, because only a fraction of the acid dissociates according to its Ka values.

In practice, we start by understanding the reaction and its equilibrium constant. Then, we use this data to construct an equilibrium expression. For the first dissociation of H2Se, this involves setting up Ka1 = [H+][HSe-]/[H2Se]. As Ka1 is the only concerning constant due to its magnitude, we can assume all H+ comes from the first ionization step. From here, we can solve for [H+] using stoichiometry and the equation, making simplifications if appropriate, as was done in the exercise with the first ionization step resulting in equal concentrations of H+ and HSe-.

pH Calculation

pH calculation allows us to express the concentration of H+ ions in a more intuitive logarithmic scale, where a decrease by one pH unit represents a tenfold increase in acidity. The pH is calculated using the formula pH = -log[H+].

Going back to our example with H2Se, after determining the H+ ion concentration from the first ionization (0.0057 M), we simply take the negative logarithm of this value to find the pH. This calculation reveals the power of the H+ ion concentration to affect the acidity of the solution. In this exercise, the calculated pH of 2.24 falls into the acidic range (pH < 7), which matches our expectation for a solution of a moderately strong acid like H2Se.

Remember, the pH scale is inverse to H+ concentration—a lower pH means a higher concentration of H+ ions, thus a more acidic solution. Also, very small changes in H+ concentration can result in significant pH changes, underscoring the importance of precise calculations.