A saturated solution of AgCl is treated with solid NaCl until the [Cl−] is 0.46 M. What will be the resulting [Ag+] in solution? Enter your answer in scientific notation.×10MWhat percent of Ag+ remains in solution at this point?

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Explanation:

Ksp of AgCl = 1.8*10^-10

Ksp = [Ag+] [Cl-]

[Ag+] [Cl-] = 1.8*10^-10

Because [Ag+] = [Cl-] ,

[Ag+]² = 1.8*10^-10

[Ag+] = √((1.8*10^-10)_

[Ag+] = 1.35*10^-5M

That is [Ag+] in the original saturated solution.

NaCl is added to the saturated AgCl solution:

If Cl- = 0.72M , equation becomes

[Ag+] [ 0.72] = 1.8*10^-10

[Ag+] = (1.8*10^-10)/0.72

[Ag+] = 2.5*10^-10M

% Ag+ remaining in solution =(2.5*10^-10) / { (1.35*10^-5) * 100

% Ag+ remaining = 1.85*10^-3% or 0.00185%

Note that an approximation has been made here: [Cl-] is 0.72M + the unprecipitated Cl- from the AgCl. Because this latter is an infinitely small amount compared to 0.72M , it is quite safe to make this approximation without any effect on the answer.

220

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.

they are all solids.