Answer :
1. Solubility in Pure Water: 1.4 × 10⁻⁷ mol/L 2. Solubility in 0.0060 M Na₂CrO₄ Solution: 3.3 × 10⁻¹² mol/L
To calculate the solubility of PbCrO₄ in pure water and in a 0.0060 M Na₂CrO₄ solution at 25 °C, we'll use the solubility product constant (Ksp) and the concept of molar solubility.
1. Solubility in Pure Water:
In pure water, PbCrO₄ dissociates into Pb²⁺ and CrO₄²⁻ ions. The solubility product constant equation is:
Ksp = [Pb²⁺][CrO₄²⁻]
Let's assume that the solubility of PbCrO₄ in pure water is x (in mol/L). Since the stoichiometric ratio between PbCrO₄ and Pb²⁺ is 1:1, the concentration of Pb²⁺ will also be x (in mol/L). Similarly, the concentration of CrO₄²⁻ will also be x (in mol/L).
Substituting these values into the Ksp equation, we get:
2.00 × 10⁻¹⁴ = (x)(x)
Simplifying the equation:
x² = 2.00 × 10⁻¹⁴
Taking the square root of both sides:
x = √(2.00 × 10⁻¹⁴)
Calculating the value:
x ≈ 1.4 × 10⁻⁷ mol/L
Therefore, the solubility of PbCrO₄ in pure water at 25 °C is approximately 1.4 × 10⁻⁷ mol/L.
2. Solubility in 0.0060 M Na₂CrO₄ Solution:
In the presence of the Na₂CrO₄ solution, the concentration of CrO₄²⁻ ions is no longer negligible. We can calculate the common ion effect on the solubility of PbCrO₄.
Assuming the solubility of PbCrO₄ in the Na₂CrO₄ solution is y (in mol/L), the concentration of CrO₄²⁻ ions will be 0.0060 M (given). The concentration of Pb²⁺ ions will still be y (in mol/L).
Applying the Ksp equation:
Ksp = [Pb²⁺][CrO₄²⁻]
2.00 × 10⁻¹⁴ = (y)(0.0060)
Simplifying the equation:
y = (2.00 × 10⁻¹⁴) / 0.0060
Calculating the value:
y ≈ 3.3 × 10⁻¹² mol/L
Therefore, the solubility of PbCrO₄ in a 0.0060 M Na₂CrO₄ solution at 25 °C is approximately 3.3 × 10⁻¹² mol/L.
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The complete question is:
Calculate the solubility at 25 °C of PbCrO₄ in pure water.
Calculate the solubility at 25 °C of PbCrO₄ in a 0.0060 M Na₂CrO₄ solution.
Please note that the given solubility constant, Ksp, for PbCrO₄ is 2.00 × 10⁻¹⁴.
