Question

Calculate the \(\mathrm{pH}\) and \(\left[\mathrm{S}^{2-}\right]\) in a \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}\) solution. Assume \(K_{\mathrm{a}_{1}}=1.0 \times 10^{-7} ; K_{\mathrm{a}_{2}}=1.0 \times 10^{-19}\)

Short answer

In summary, for a 0.10 M H₂S solution with Ka₁ = 1.0 x 10⁻⁷ and Ka₂ = 1.0 x 10⁻¹⁹, the pH is 4, and the concentration of S²⁻ ions is 1.0 x 10⁻¹⁹ M.

Step-by-step solution

Write the ionization reactions and dissociation constants

For the two-step ionization of H₂S, we have the following reactions and dissociation constants: 1. First ionization: \(H_2S \rightleftharpoons H^+ + HS^-\) Dissociation constant Ka₁: \[K_{a_{1}} = \frac{[H^+][HS^-]}{[H_{2}S]} = 1.0×10^{-7}\] 2. Second ionization: \(HS^- \rightleftharpoons H^+ + S^{2-}\) Dissociation constant Ka₂: \[K_{a_{2}} = \frac{[H^+][S^{2-}]}{[HS^-]} = 1.0×10^{-19}\]

Write the ionization expressions

Let x mol of H₂S ionizes. So, the concentration of each species after the first ionization can be expressed as follows: \[ [H^+] = x \] \[ [HS^-] = x \] \[ [H_2S] = 0.10 - x \] Now, substitute these expressions in Ka₁: \[1.0×10^{-7} = \frac{x \cdot x}{0.10 - x}\]

Solve for x and calculate the pH

As Ka₁ is a weak acid dissociation constant, the value of x will be significantly smaller than the initial concentration (0.10) of H₂S. Therefore, we can assume \(x ≪ 0.10\). This allows us to simplify the equation for x: \[1.0×10^{-7} ≈ \frac{x^2}{0.10}\] Solve for x: \[x ≈ \sqrt{1.0×10^{-7} \cdot 0.10} = 1.0×10^{-4}\] So, the concentration of H⁺ ions is approximately 1.0 x 10⁻⁴ M. Now, the pH can be calculated: \[pH = -\log{[H^+]} = -\log{(1.0×10^{-4})} = 4\]

Calculate the concentration of S²⁻ using Ka₂

Knowing the concentration of the H⁺ ion and the first ionization product (HS⁻), we can now proceed to calculate the concentration of the sulfide ion (S²⁻) using Ka₂: \[1.0×10^{-19} = \frac{(1.0×10^{-4}) [S^{2-}]}{x}\] Substitute x = 1.0 x 10⁻⁴ \[1.0×10^{-19} = \frac{(1.0×10^{-4}) [S^{2-}]}{1.0×10^{-4}}\] Solve for the concentration of S²⁻: \[[S^{2-}] = 1.0×10^{-19}\] So, the concentration of sulfide ions (S²⁻) is 1.0 x 10⁻¹⁹ M. To summarize, the pH of the 0.10 M H₂S solution is 4, and the concentration of S²⁻ ions is 1.0 x 10⁻¹⁹ M.

Key concepts

Ionization Reactions

Ionization reactions occur when a molecule dissociates into ions. This is a common process in solutions containing acids and bases. For our exercise, the ionization of hydrosulfuric acid is analyzed in two steps. In the first ionization reaction, H₂S dissociates into H⁺ ions and HS⁻ ions. The relevant reaction is:
- \( H_2S \rightleftharpoons H^+ + HS^- \) Each dissociation reaction has an associated equilibrium constant, known as the dissociation constant.
The second ionization is:
- \( HS^- \rightleftharpoons H^+ + S^{2-} \) Understanding these ionization reactions helps us calculate the concentrations of ions in a solution, which in turn allows us to determine the pH and other properties of the solution.

Dissociation Constants

Dissociation constants, represented by \( K_a \), are crucial for understanding how well an acid dissociates in water. They give us a measure of the acid's strength and the equilibrium position of its ionization reaction. For our problem, there are two dissociation constants given for H₂S.
* The first dissociation constant (\( K_{a1} \)) is \( 1.0 \times 10^{-7} \).* It relates to the equilibrium of the reaction \( H_2S \rightleftharpoons H^+ + HS^- \).* The second dissociation constant (\( K_{a2} \)) is \( 1.0 \times 10^{-19} \).* This constant applies to the next equilibrium: \( HS^- \rightleftharpoons H^+ + S^{2-} \).
These constants allow us to write expressions that predict the concentration of products and reactants at equilibrium. Using these expressions, we can solve for unknowns such as ion concentrations and eventually the pH of the solution.

Weak Acid Dissociation

Weak acids, like H₂S, do not completely ionize in solution. This partial ionization is what we are dealing with when calculating pH in this exercise. In the case of H₂S, the first dissociation has a low \( K_a \) value of \( 1.0 \times 10^{-7} \), indicating modest ionization. The second dissociation is even weaker with a \( K_a \) of \( 1.0 \times 10^{-19} \), signifying negligible ionization under normal conditions. This means:
- Most H₂S remains as un-ionized molecules in solution.- Only a small portion produces H⁺ and HS⁻ ions.- Even fewer HS⁻ ions produce S²⁻ in the second dissociation.
Understanding weak acid behavior is essential to correctly determine pH and concentrations of different ions in solutions.

Sulfide Ion Concentration

The sulfide ion (S²⁻) emerges from the second step of H₂S ionization. Due to the extremely small \( K_{a2} \) for the second dissociation, the concentration of S²⁻ ions in solution is very low. Here’s why:
- The ionization equilibrium strongly favors the reactants, meaning very few S²⁻ are produced.- In our solution, even after calculating, the concentration of \([S^{2-}]\) is observed to be \( 1.0 \times 10^{-19} \) M.
While the concentration of sulfide ions is generally negligible, it is still essential to consider it in calculations for thoroughness and accuracy, particularly in chemical environments where trace ions can have significant effects.