Calculate the solubility of AgCl in:
Ksp = 1.6x10^-10
a) 100 ml of 4.00 x 10^-3 M Calcium chloride
b) 100 ml of 4.00 x 10^-3 M Calcium Nitrate
all i know is 1.6x10^-10=[Ag+][Cl-]
which is 1.6x10^-10=x^2
x=1.3x10^-5, giving the M of Ag and Cl. Don't know how to go further
What you know to do, and very well I might add, is to calculate Ag and Cl in a saturated solution of AgCl.
These two problems are slightly different. Basically they illustate the common ion effect. The first one has CaCl2 in the solution (the common ion is Cl^-) and the second one is calcium nitrate (no common ion). Use your method to calculate the solubility in Ca(NO3)2 since there is no common ion (technically the solubility is increased with a salt BUT I suspect you haven't covered that yet).
The first part s done this way.
AgCl(s) ==> Ag^+ + Cl^-
Ksp = (Ag^+)(Cl^-) = 1.6E-10
And
CaCl2 ==> Ca^2+ + 2Cl^-
Now substitute into the Ksp expression.
(Ag^+) = x
(Cl^-) = x from the AgCl and 0.008M for a total of x+0.008. (Where did the 0.008 come from? That is 0.004 M CaCl2 so the chloride is just twice that.) Solve for x and that will be the molarity. That is moles/L; you want grams/100 mL. So moles/L x 0.1 L (from the 100 mL) gives you moles/100 mL and that x molar mass = grams AgCl.
.000032
3 pi
Well, let me help you with that! Now that you have the molar concentration of Ag+ and Cl-, you can use that to calculate the solubility. Remember that the solubility is the molar concentration of the solute in a saturated solution at a given temperature.
a) In 100 ml of 4.00 x 10^-3 M Calcium chloride, you have 4.00 x 10^-3 moles of Cl- ions. Since 1 mole of AgCl produces 1 mole of Cl-, the solubility of AgCl in this solution is also 4.00 x 10^-3 M.
b) Similarly, in 100 ml of 4.00 x 10^-3 M Calcium Nitrate, you have 4.00 x 10^-3 moles of NO3- ions. However, Calcium Nitrate does not provide any Cl- ions, so the solubility of AgCl in this solution would still be 0 M.
So, to summarize:
a) The solubility of AgCl in 100 ml of 4.00 x 10^-3 M Calcium chloride is 4.00 x 10^-3 M.
b) The solubility of AgCl in 100 ml of 4.00 x 10^-3 M Calcium Nitrate is 0 M.
Hope that clears things up, or maybe I just made things more confusing!
To calculate the solubility of AgCl in the given solutions, you need to consider the common ion effect. The presence of other ions can affect the solubility of a compound, so we need to take that into account.
a) 100 ml of 4.00 x 10^-3 M Calcium chloride (CaCl2):
We need to determine the concentration of Cl- ion provided by CaCl2. Since CaCl2 dissociates into Ca2+ and 2 Cl- ions, the concentration of Cl- is twice the concentration of CaCl2.
Concentration of Cl- = 2 * 4.00 x 10^-3 M = 8.00 x 10^-3 M
Now, we can set up the equilibrium expression with AgCl:
Ksp = [Ag+][Cl-]
Ksp = (x)(8.00 x 10^-3) [Since AgCl dissociates into Ag+ and Cl- ions, we can assume x M concentration for Ag+]
1.6 x 10^-10 = 8.00 x 10^-3 * x
x = (1.6 x 10^-10) / (8.00 x 10^-3)
x = 2 x 10^-8 M
So, the solubility of AgCl in 100 ml of 4.00 x 10^-3 M Calcium chloride is 2 x 10^-8 M.
b) 100 ml of 4.00 x 10^-3 M Calcium Nitrate (Ca(NO3)2):
We need to determine the concentration of NO3- ion provided by Ca(NO3)2. Since Ca(NO3)2 dissociates into Ca2+ and 2 NO3- ions, the concentration of NO3- is twice the concentration of Ca(NO3)2.
Concentration of NO3- = 2 * 4.00 x 10^-3 M = 8.00 x 10^-3 M
Since there are no Cl- ions provided by Ca(NO3)2 solution, the concentration of Cl- remains zero.
Using the equilibrium expression: Ksp = [Ag+][Cl-]
Ksp = (x)(0) [Since there are no Cl- ions provided by Ca(NO3)2, we can assume the concentration of Cl- as zero]
1.6 x 10^-10 = 0 * x
x = 0
So, the solubility of AgCl in 100 ml of 4.00 x 10^-3 M Calcium Nitrate is zero.