A solution was prepared such that the concentrations of each of the ions Br^-, CO₃^{2 -}, CN^-, I^-, and PO₄^{3 -} were 1.00×10^{- 4} M . If Ag⁺ ions are then added to this solution, which of the following precipitates would form first ?

Question

Quizplus · Accepted Answer

To determine which precipitate will form first, we need to compare the solubility product (Ksp) values for each compound. The compound with the lowest Ksp value will form a precipitate first.

Using the given concentrations and the balanced chemical equations, we can write the ionic products for each compound:

AgBr: [Ag+][Br-] = (x)(1.00×10^-4)
Ag2S/CuS: [Ag+][S2-] = (x)(1.00×10^-4)
AgCN: [Ag+][CN-] = (x)(1.00×10^-4)
AgI: [Ag+][I-] = (x)(1.00×10^-4)
Ag3PO4: [Ag+][PO43-] = (x)(1.00×10^-4)

where x is the molar solubility of the precipitate.

Using the given Ksp values, we can write the equilibrium expressions and solve for x:

AgBr: Ksp = [Ag+][Br-] = x^2
x = sqrt(Ksp) = sqrt(8.5×10^-17) = 9.2×10^-9 M

Ag2S/CuS: Ksp = [Ag+][S2-] = x^2
x = sqrt(Ksp) = sqrt(1.0×10^-51) = 1.0×10^-26 M

AgCN: Ksp = [Ag+][CN-] = x^2
x = sqrt(Ksp) = sqrt(1.2×10^-16) = 1.1×10^-8 M

AgI: Ksp = [Ag+][I-] = x^2
x = sqrt(Ksp) = sqrt(8.5×10^-17) = 9.2×10^-9 M

Ag3PO4: Ksp = [Ag+][PO43-] = x^4
x = fourth root of(Ksp) = fourth root of(1.0×10^-51) = 1.0×10^-13 M

The compound with the lowest molar solubility, and therefore the one that will form a precipitate first, is AgI. Therefore, the answer is choice D.

A Solution Was Prepared Such That the Concentrations of Each