Tap water that contains Cl− at a concentration
of 50 ppm is used to prepare a 0.100 mol/L
AgNO3 solution. Does a precipitate form? How much?
I don't know how to find out how much precipitate will form but a precipitate will form based on my calculations
50 ppm 0.0014 mol/l
0.1 Mol/l = AgNO3
Ag+ = 0.1 Mol/l
Cl-=0.0014 mol/l
Q= [ag+][Cl-]
this is the rest of the calculations
Q= 1.4x10^-4
ksp= 1.77x10^-10
Q>ksp so precipitation will form
You're ok as far as you went.
Who ever made up this problem forgot to tell us how much solution is being prepared. It makes sense, doesn't it, that we can get more AgCl if we made up 5L than if we made up 1L. So I guess we assume we make 1 L solution.
........Ag^+ + Cl^- ==> AgCl
I......0.1mol.0.00141.....0
C..-0.00141.-0.00141....0.00141
E...............0........0.00141
So we will have 0.00141 mols AgCl ppt and that x molar mass AgCl gives you the grams AgCl. I assume it's obvious that Cl^- is the limiting reagent.
To determine whether a precipitate will form in this case and how much, you need to compare the concentration product, known as the reaction quotient (Q), to the solubility product constant (Ksp) for silver chloride (AgCl).
The equation for the formation of the precipitate is:
Ag+(aq) + Cl-(aq) ⇌ AgCl(s)
First, you have calculated the concentration of Ag+ to be 0.1 mol/L because you are preparing a 0.100 mol/L AgNO3 solution.
Secondly, you have calculated the concentration of Cl- in the tap water, which is 50 ppm or 0.0014 mol/L.
Now, you can substitute these values into the reaction quotient expression:
Q = [Ag+][Cl-]
Since the concentrations are known:
Q = (0.1 mol/L) * (0.0014 mol/L)
Q = 0.00014 mol^2/L^2
Next, you need to compare this with the solubility product constant (Ksp) for silver chloride, which is 1.6 x 10^-10 at 25 °C.
If Q is greater than Ksp, a precipitate will form. If Q is less than Ksp, no precipitate will form.
In this case:
Q = 0.00014 mol^2/L^2
Ksp = 1.6 x 10^-10
Since Q is greater than Ksp, a precipitate will form. The amount of precipitate formed depends on how much Ag+ and Cl- ions are present in excess of the solubility product.
To determine the amount of precipitate formed, you need to find the limiting reactant. From the balanced equation, you can see that the molar ratio between Ag+ and Cl- is 1:1. Since the concentration of Cl- is less than Ag+, Cl- is the limiting reactant.
The molar ratio allows you to determine the moles of AgCl precipitate formed. In this case, it will be equal to the moles of Cl-:
moles of AgCl = moles of Cl- = 0.0014 mol/L
To find the mass of AgCl precipitate formed, you need to multiply the moles of AgCl by its molar mass, which is 143.32 g/mol.
mass of AgCl = moles of AgCl * molar mass of AgCl
mass of AgCl = 0.0014 mol/L * 143.32 g/mol
mass of AgCl = 0.2001 g
Therefore, in this scenario, a precipitate of silver chloride (AgCl) will form, and the mass of the precipitate will be approximately 0.2001 grams.