Determine the molar solubility of BiPO₄ in a solution containing 0.358 M K₃PO₄.The Ksp for BiPO₄is 1.3 × 10^-23. A) 5.7 × 10^-15 B) 3.6 × 10^-23 C) 1.3 × 10^-13 D) 1.5 × 10^-8 E) 8.3 × 10^-18

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There are a few steps to solve this problem. First, write out the balanced equation for the dissolution of BiPO₄: BiPO₄(s) ⇌ BiPO₄(aq) Next, write out the expression for the solubility product (Ksp): Ksp = [BiPO₄(aq)] Since we do not know the molar solubility of BiPO₄, we can define it as "s": Ksp = s2 We also know that the solution contains 0.358 M K₃PO₄. Using the stoichiometry of the balanced equation, we can see that the concentration of BiPO₄ in the solution will also be "s": [BiPO₄(aq)] = s [K₃PO₄(aq)] = 0.358 M [PO₄(aq)] = 0 (since it is a spectator ion) Now we can plug these values into the expression for Ksp: 1.3 × 10^-23 = s2 Solving for "s", we get: s = 3.6 × 10^-12 This is the molar solubility of BiPO₄ in the solution. However, the question asks for the molar solubility in the format of scientific notation with only one digit to the left of the decimal point. Therefore, the answer is: B) 3.6 × 10^-23

Determine the Molar Solubility of BiPO4 in a Solution Containing

Determine the Molar Solubility of BiPO₄ in a Solution Containing