A certain municipal water sample contains 46.1 mg SO42-/L and 30.6 mg Cl -/L. How many drops (1 drop = 0.05 mL) of 0.0010 M AgNO3 must be added to 1.00 L of this water to just produce a precipitate?

What will this precipitate be?

how do i do this problem?

agcl ksp = 1.8 e-10

ag2so4 ksp = 1.5 e10-5

Ksp AgCl = (Ag^+)(Cl^-) = 1.8 x 10^-10

(Cl^-) = 30.6 mg from the problem.
moles Cl = 30.6 x 10^-3 g/35.45 = ??moles and since that is in 1 L of the water sample, that is the molarity.
Plug Cl^- (in moles/L) into Ksp and solve for (Ag^+) required for AgCl to ppt.
Then M Ag^+ x 1000 mL = moles Ag^+.

Since moles = M x L. You know moles Ag^+ needed and you know M of the AgNO3 solution, calculate L and convert that to drops. Post your work if you get stuck.
You work the Ag2SO4 the same way.

I did the moles of Cl found/1 L = .00863.

1.8e-10 / .00863 = 2.1e-8 * 1 L = 2.1e-8 moles

.0010 / 2.1e-8 = 47619 L*1000 mL / .5

but the answer comes out wrong=/

I think (Cl^-) = 0.0306/35.45 = 8.63 x 10^-4 moles which is a factor of 10 different from your value. That will make (Ag^+) needed to ppt AgCl be (1.8 x 10^-10/8.63 x 10^-4 = 2.085 x 10^-7 M instead of your value. (I know that's too many significant figures but I carry too many and round at the end). From here on out, I don't follow your work at all.

Since this is 2.085 x 10^-7 moles/L and we have 1 L of solution, we have 2.085 x 10^-7 moles in the 1 L. Then L = (moles/M) = 2.085 x 10^-7/0.001) = 2.085 x 10^-4 L or 2.085 x 10^-1 mL = 0.2085 mL. Check my work
Finally. 0.2085 mL x (1 drop/0.05 mL) = 4.17 drops which I would round to 5 drops. (4.17 rounds to 4 drops to 1 significant figure which is all we are allowed based on the 0.05 mL/drop but 4 drops will be too small to cause a ppt and we must take 5 drops). So 5 drops it is. Check my arithmetic.

To solve this problem, we need to use the concept of solubility product constant (Ksp). The equation for the reaction between AgNO3 and Cl- is:

AgNO3(aq) + Cl-(aq) ↔ AgCl(s) + NO3-(aq)

The Ksp expression for this reaction is:

Ksp = [Ag+][Cl-]

Given that the Ksp of AgCl is 1.8 x 10^-10, our objective is to find the point at which the concentration of Ag+ and Cl- ions exceeds this value, leading to the formation of a precipitate.

1. Start by calculating the concentration of Cl- ions in the water sample as follows:

Concentration of Cl- = 30.6 mg/L = 30.6 mg/dm^3 = 30.6 mg/dm^3 / (10^3 mg/g) / (10^-3 dm^3/L) = 0.0306 g/L

2. Next, convert the g/L concentration to mol/L (M):

Molar mass of Cl- = 35.45 g/mol

Concentration of Cl- in mol/L (M) = 0.0306 g/L / 35.45 g/mol = 0.000862 M

3. Since the stoichiometry of the reaction is 1:1, the concentration of Ag+ ions is also 0.000862 M.

4. For precipitation to occur, the concentration of Ag+ ions must equal or exceed the Ksp value. Therefore, we set up the following equation:

[Ag+][Cl-] = Ksp
(0.000862 M)(0.000862 M) = 1.8 x 10^-10

5. Solve the equation for [Ag+]:

[Ag+] = sqrt(Ksp / [Cl-])
[Ag+] = sqrt(1.8 x 10^-10 / 0.000862)

[Ag+] ≈ 6.89 x 10^-6 M

6. Now, we need to determine the number of moles of AgNO3 required to achieve this concentration of Ag+ in the 1.00 L of water. We can use the equation:

moles of AgNO3 = volume of AgNO3 (in L) x molarity of AgNO3

Let's assume the volume of AgNO3 required is V mL. Converting V mL to L:

V L = V mL x (1 drop / 20 drops) x (0.05 mL / 1 drop) x (1 L / 1000 mL)
V L = V/4000

moles of AgNO3 = V/4000 * 0.0010 M

7. Finally, equate the moles of AgNO3 to the moles derived from the concentration of Ag+ calculated above:

moles of AgNO3 = (0.000862 M)(1.00 L)
V/4000 * 0.0010 M = 0.000862 mol

8. Solve for V:

V = (0.000862 mol) / (0.0010 M) * (4000/1)
V ≈ 3450 drops

Therefore, approximately 3450 drops (1 drop = 0.05 mL) of 0.0010 M AgNO3 must be added to the 1.00 L of water to just produce a precipitate.

The precipitate formed will be AgCl (silver chloride), as predicted by the balanced equation.