Calculate the concentration of silver ions at equilibrium when #"0.5 g"# of #"AgCl"# is placed in #"100 mL"# of water?
Ksp = #1.8 *10^-10#
Ksp =
1 Answer
Here's what I got.
Explanation:
The trick here is to realize that because silver chloride has a solubility of about
This implies that a lot of the silver chloride that you're adding will remain undissolved, meaning that it will not affect the equilibrium concentrations of the silver(I) cations and of the chloride anions.
So, when silver chloride is placed in water, the following equilibrium is established.
#"AgCl"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#
Notice that for every
By definition, the solubility product constant for this equilibrium,
#K_(sp) = ["Ag"^(+)] * ["Cl"^(-)]#
If you take
#K_(sp) = 1.8 * 10^(-10)#
you can say that you have
#K_(sp) = s * s#
#K_(sp) = s^2#
This will get you
#s = sqrt(K_(sp))#
#s = sqrt(1.8 * 10^(-10)) = 1.34 * 10^(-5)#
This means that at the temperature used here, a saturated solution of silver chloride will have
#["Ag"^(+)] = 1.34 * 10^(-5)color(white)(.)"M"#
#["Cl"^(-)] = 1.34 * 10^(-5)color(white)(.)"M"#
In other words, you can only hope to dissolve
Consequently, you can say that
#100 color(red)(cancel(color(black)("mL solution"))) * (1.34 * 10^(-5)color(white)(.)"moles AgCl")/(10^3color(red)(cancel(color(black)("mL solution")))) = 1.34 * 10^(-6)color(white)(.)"moles AgCl"#
Use the molar mass of silver chloride to convert this to grams
#1.34 * 10^(-6) color(red)(cancel(color(black)("moles AgCl"))) * "143.32 g"/(1color(red)(cancel(color(black)("mole AgCl")))) = 1.92 * 10^(-4)color(white)(.)"g"#
Therefore, you can say that if you add
#1.92 * 10^(-4) color(white)(.)"g" = 192color(white)(.)mu"g"#
of silver chloride will actually dissociate to produce silver(I) cations and chloride anions. The rest will remain undissolved.