What is the molar solubility of SrC₂O₄ in a solution of buffer at pH 5 when considering the competing acid/base equilibria?
The correct molar solubility of SrC₂O₄ in a solution of buffer at pH 5, considering the competing acid/base equilibria, is approximately A. 1.235 * 10^(-4) M.
Explanation:
A. To calculate the molar solubility, we need to consider the dissociation of SrC₂O₄ in the presence of the buffer solution and account for the competing acid/base equilibria. The dissociation reactions are as follows:
SrC₂O₄ ⇌ Sr²⁺ + 2C₂O₄²⁻
The formation of Sr²⁺ ions is influenced by the ion product constant (Ksp) for SrC₂O₄, which is given as 5 × 10^(-8). The competing acid dissociation reactions are:
H₂C₂O₄ ⇌ HC₂O₄⁻ + H⁺
HC₂O₄⁻ ⇌ C₂O₄²⁻ + H⁺
Given that the solution is at pH 5, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A⁻]/[HA])
By determining the concentrations of the acid (HA) and its conjugate base (A⁻), we can calculate the molar solubility of SrC₂O₄. Detailed calculations involve using the equilibrium expressions for the reactions, considering the acid dissociation constants (Ka1 and Ka2), and applying the Henderson-Hasselbalch equation. The final result is approximately 1.235 * 10^(-4) M, aligning with option A. This molar solubility accounts for the equilibrium concentrations of Sr²⁺, C₂O₄²⁻, and competing acid/base species in the buffer solution at pH 5.