An aqueous solution contains 0.010 M bromide ion (Br^-)and 0.010 M iodide ion (I^-).If silver ions (Ag⁺)are added to the solution until silver bromide just begins to precipitate,what are the concentrations of silver ions and iodide ions? (K_sp of AgBr = 5.4 × 10^-13 and K_sp of AgI = 8.5 × 10^-17) A) [Ag⁺] = 5.4 × 10^-11 M,[I^-] = 1.0 × 10^-2 M B) [Ag⁺] = 8.5 × 10^-15 M,[I^-] = 1.0 × 10^-2 M C) [Ag⁺] = 5.4 × 10^-11 M,[I^-] = 1.6 × 10^-6 M D) [Ag⁺] = 8.5 × 10^-15 M,[I^-] = 6.4 × 10¹ M E) [Ag⁺] = 8.5 × 10^-15 M,[I^-] = 1.6 × 10^-6 M

Question

Quizplus · Accepted Answer

The precipitation reaction is:

Ag+ + Br- → AgBr (Ksp = 5.4 × 10^-13)

At the point where AgBr just begins to precipitate, the concentrations of Ag+ and Br- in the solution are equal to the solubility product (Ksp) of AgBr.

So, [Ag+] = [Br-] = sqrt(Ksp(AgBr)) = sqrt(5.4 × 10^-13) = 7.35 × 10^-7 M

The remaining iodide ion concentration is given by:

[I-] = [I-]initial - [Ag+] = 0.010 - 7.35 × 10^-7 = 0.010 - 0.000000735 = 0.009999265 M

Now, check the solubility product of AgI:

Ag+ + I- → AgI (Ksp = 8.5 × 10^-17)

If [I-] = 0.009999265 M, then:

Ksp(AgI) = [Ag+][I-] = (7.35 × 10^-7)(0.009999265) = 7.349 × 10^-12

This value is much smaller than the Ksp of AgI, so there is no precipitation of AgI. Therefore, the concentration of iodide ions remains at 0.009999265 M.

So, the final concentrations are:

[Ag+] = 7.35 × 10^-7 M
[I-] = 0.009999265 M

Therefore, the best choice is C.

An Aqueous Solution Contains 0