Determine the molar solubility of MX (Ksp=4.2x10-8) in 0.191 M NaCN. The metal ion M+ forms M(CN)2- (Kf=3.1x1010) in the presence of CN- ion.
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To determine the molar solubility of MX in the presence of NaCN, we need to consider the equilibrium between the dissolution and precipitation of MX. We also need to take into account the formation of the complex ion M(CN)2-.
Let's break down the process step by step:
Step 1: Write the balanced equation for the dissolution of MX in water:
MX ↔ M+ + X-
Step 2: Write the equilibrium expression for the dissolution of MX:
Ksp = [M+][X-]
Step 3: Consider the presence of CN- ions from NaCN, which will react with M+ ions to form the complex ion M(CN)2-. The formation constant for this reaction is given as Kf = 3.1x10^10.
Step 4: Write the balanced equation for the formation of the complex ion M(CN)2-:
M+ + 2CN- ↔ M(CN)2-
Step 5: Write the equilibrium expression for the formation of the complex ion M(CN)2-:
Kf = [M(CN)2-]/([M+][CN-]^2)
Step 6: Since the concentration of CN- is given as 0.191 M, we can assume that the complex ion formation is largely complete, and its concentration remains constant.
Step 7: Substitute the value of [CN-] into the equilibrium expression for Kf to simplify it, and then rearrange to solve for [M(CN)2-]:
[M(CN)2-] = Kf * [M+] * [CN-]^2
Step 8: Substitute the value of [M(CN)2-] into the equilibrium expression for Ksp to get an expression that can be solved for [M+]:
Ksp = [M+] * [X-] = [M+] * ([CN-]^2 / Kf)
Step 9: Substitute the known values into the equation and solve for [M+]:
4.2x10^-8 = [M+] * (0.191 M)^2 / (3.1x10^10)
Solving this equation will give you the concentration of [M+], which represents the molar solubility of MX in the presence of NaCN.