Calculate the solubility at 25 °C of BaCrO4 in pure water and in a 0.0180M BaCI2 solution.

Question

Calculate the solubility at 25 °C of BaCrO₄ in pure water and in a 0.0180M BaCI₂ solution.

BaCrO₄ K_sp= 1.17 x 10^-10

Round both of your answers to 2 significant digits.

solubility in pure water: ____g/L

solubility in 0.0180 M BaCI₂ Solution: ____g/L

Answer 1

BaCrO₄(s) <==> Ba²⁺(aq) + CrO₄^2-(aq)

Ksp = [Ba²⁺][CrO₄^2-]

1.17x10^-10 = (x)(x) = x²

x = [BaCrO₄] = 1.1x10^-5 mol/L

To convert to g/L use the molar mass of BaCrO₄ (253 g / mol)

1.1x10^-5 mol/L x 253 g / mol = 2.7x10^-3 g/L in pure water

To find solubility in 0.0180 M BaCl₂ solution, we look at the common ion effect. In this case the common ion is Ba²⁺. We can approximate the total [Ba²⁺] as coming from the BaCl₂ as it is much greater than that supplied by the BaCrO₄ (1.1x10^-5 M). According to LeChatelier, we can expect the solubility of BaCrO₄ to DECREASE compared to that in pure water.

Ksp = [Ba²⁺][CrO₄^2-]

1.17x10^-10 = [0.0180][CrO₄^2-]

[CrO₄^2-] = 6.5x10^-9 M

converting to g/L we have...

6.5x10^-9 mol/L x 253 g/ mol = 1.6x10^-6 g/L = solubility in 0.0180 M BaCl₂