Question

Calculate the $\mathrm{pH}$ and $\left[{\mathrm{S}}^{2-}\right]$ in a $0.10-M{\mathrm{H}}_2\mathrm{S}$ solution. Assume $K_{{\mathrm{a}}_1}=1.0\times {10}^{-7};K_{{\mathrm{a}}_2}=1.0\times {10}^{-19}$

Short answer

In a 0.10 M H₂S solution with given ionization constants $K_{a_1}=1.0\times {10}^{-7}$ and $K_{a_2}=1.0\times {10}^{-19}$, the pH is calculated to be 4 and the concentration of S²⁻ ions is $1.0\times {10}^{-15}$ M.

Step-by-step solution

Write Equilibrium Equations

For ionization constants, we need to set up the equilibrium expressions for H₂S: Equation 1: $H_{2}S\rightleftharpoons H^{+}+H{S}^{-}$ with equilibrium constant, Ka₁ Equation 2: $H{S}^{-}\rightleftharpoons H^{+}+S^{2-}$ with equilibrium constant, Ka₂

Write Expressions for Ka₁ and Ka₂

Now we will write the expressions for the ionization constants Ka₁ and Ka₂: $K_{a_1}=\frac{[H^{+}][H{S}^{-}]}{[H_{2}S]}$ and $K_{a_2}=\frac{[H^{+}][S^{2-}]}{[H{S}^{-}]}$

Approximate the Concentrations

Since Ka₂ is much smaller than Ka₁, we can assume for the time being that the majority of hydrogen ions come from the first ionization: Let x be the concentration of H⁺ ions, so [H⁺] = x. For the first equilibrium, [HS⁻] ≈ x and [H₂S] ≈ 0.10 M. We can ignore the contribution from the second ionization in this step.

Solve for x (H⁺ Concentration)

Using the expression for Ka₁ and the given value of Ka₁, we can solve for x: $1.0\times {10}^{-7}=\frac{x^2}{0.10}$ $x^2=1.0\times {10}^{-8}$ $x=\sqrt{1.0\times {10}^{-8}}=1.0\times {10}^{-4}$ Now we have the H⁺ ion concentration, and we can calculate the pH.

Calculate the pH

Using the H⁺ ion concentration, we can determine the pH. pH = -log10([H⁺]) = -log10(1.0 × 10⁻⁴) = 4 Now, we will find the concentration of S²⁻ ions.

Solve for [S²⁻] Concentration

Using the second equilibrium equation and the value of Ka₂, we can find the concentration of S²⁻ ions, [S²⁻], by substituting [H⁺]=1.0 x 10⁻⁴ and assuming that [HS⁻] ≈ x which we found in the previous steps: $K_{a_2}=\frac{[H^{+}][S^{2-}]}{[H{S}^{-}]}$ $1.0\times {10}^{-19}=\frac{(1.0\times {10}^{-4})[S^{2-}]}{1.0\times {10}^{-4}}$ $$S^{2-}=1.0\times {10}^{-15}$$ So, the pH of the 0.10 M H₂S solution is 4 and the concentration of S²⁻ ions is $1.0\times {10}^{-15}$ M.

Key concepts

H₂S ionization

Understanding the ionization process of hydrogen sulfide (H₂S) is essential for calculating its pH and other relevant concentrations in solution. H₂S is a weak diprotic acid, meaning it can donate two protons (H⁺ ions) but does so in two separate stages.

• First ionization: In the initial step, H₂S ionizes to form hydrosulfide ions (HS⁻) and hydrogen ions (H⁺). This process is represented by the equation: $$H_{2}Sightleftharpoons{H}^{+}+H{S}^{-}$$
• Second ionization: The hydrosulfide ion (HS⁻) can further lose a proton to form sulfide ions (S²⁻) and an additional hydrogen ion (H⁺): $$H{S}^{-}ightleftharpoons{H}^{+}+S^{2-}$$

Each stage of ionization has its own equilibrium constant. The initial ionization (first ionization) is more significant, influencing the solution's pH more profoundly than the second.

Equilibrium Expressions

Equilibrium expressions help us to quantify the extent of ionization. For each ionization step of H₂S, there's a corresponding equilibrium expression involving the concentrations of the reactants and products.

• Ka₁ Expression: For the first ionization, we use: $K_{a_1}=\frac{[H^{+}][H{S}^{-}]}{[H_{2}S]}$
• Ka₂ Expression: For the second ionization, we have: $K_{a_2}=\frac{[H^{+}][S^{2-}]}{[H{S}^{-}]}$

These expressions allow us to calculate the concentrations of ions at equilibrium. Knowing the values for biochemical or chemical systems helps predict or determine the behavior in a solution.

Ka values

Ka values are crucial for understanding how strongly an acid will ionize in a solution. The magnitude of a Ka value indicates the strength of the acid.

• Ka₁: The first ionization constant for H₂S is high relative to Ka₂, typically $1.0\times {10}^{-7}$. This indicates that while H₂S does ionize, it is still a weak acid.
• Ka₂: The second ionization constant, $1.0\times {10}^{-19}$, is significantly lower, showing that the second proton release is much less likely and far weaker.

The smaller Ka₂ implies that the formation of S²⁻ ions is minimal compared to HS⁻ ions. This affects how we simplify calculations, such as assuming the majority of H⁺ ions in solution come from the first ionization step.

Acid-Base Equilibrium

Acid-base equilibria involve reversible reactions where acids donate protons and bases accept them. In the case of H₂S, both ionization stages establish an equilibrium between reactants and products.

• First Equilibrium: The first equilibrium is more significant because it governs the solution's pH. Since Ka₁ is much larger than Ka₂, this step accounts for most H⁺ ions.
• Second Equilibrium: The second ionization provides additional insight but doesn't strongly influence pH due to its much smaller Ka value.

Understanding these equilibria allows for calculating pH and ion concentrations by setting up equations and assumptions like neglecting lesser effects. By considering only major contributors to H⁺ ion concentration, approximations simplify the complex equilibrium calculations.