If you mix 20.0 mL of 1.00 x 10-4 mol/L NaCl with 20.0 mL of 2.00 x 10-6 mol/L AgNO3 will a precipitate form ? If it does it will be AgCl, which has a Ksp of 1.80 x 10-10.

i'm not quite sure how to figure this out...

You want the solubility quotient which is simply (Ag^+)(Cl^-) in the problem, then compare the product with Ksp. If the product is larger a ppt forms. If the product is smaller, there is no ppt of AgCl. First you must realize that you are mixing solutions; therefore, they will dilute each other.

NaCl is 1.00E-4 x 20 mL/40 mL = 5E-5 M.
AgNO3 is 2.00E-6 x 20 mL/40 mL = 1E-6 M.

So the product is (5E-5)(1E-6) = 5E-11
Compare that with 1.80E-10.

Thank you so much! :)

To determine if a precipitate will form when mixing two solutions, we can compare the molar solubility of the potential precipitate (AgCl) to the concentrations of the ions in the solution (Na+ and Cl-).

Step 1: Calculate the number of moles of NaCl and AgNO3 in the solutions.
Number of moles NaCl = concentration (mol/L) x volume (L)
Number of moles NaCl = (1.00 x 10^-4 mol/L) x 0.020 L
Number of moles NaCl = 2.00 x 10^-6 mol

Number of moles AgNO3 = concentration (mol/L) x volume (L)
Number of moles AgNO3 = (2.00 x 10^-6 mol/L) x 0.020 L
Number of moles AgNO3 = 4.00 x 10^-8 mol

Step 2: Determine the limiting reactant.
The limiting reactant is the one that is completely consumed in the reaction and limits the amount of product formed. In this case, it is NaCl because it has fewer moles (2.00 x 10^-6 mol) compared to AgNO3 (4.00 x 10^-8 mol).

Step 3: Calculate the concentration of Na+ and Cl- in the final solution.
Since NaCl is the limiting reactant, all of the Na+ and Cl- ions will react to form AgCl.

Concentration of Na+ = moles of Na+ / total volume (L)
Concentration of Na+ = 2.00 x 10^-6 mol / (0.020 L + 0.020 L)
Concentration of Na+ = 5.00 x 10^-5 mol/L

Concentration of Cl- = moles of Cl- / total volume (L)
Concentration of Cl- = 2.00 x 10^-6 mol / (0.020 L + 0.020 L)
Concentration of Cl- = 5.00 x 10^-5 mol/L

Step 4: Compare the concentrations of Na+ and Cl- ions to the molar solubility of AgCl.
The molar solubility of AgCl is given as Ksp = 1.80 x 10^-10.

Since the concentrations of Na+ and Cl- (5.00 x 10^-5 mol/L) are greater than the molar solubility of AgCl (1.80 x 10^-10), a precipitate of AgCl will form.

In conclusion, mixing 20.0 mL of 1.00 x 10^-4 mol/L NaCl with 20.0 mL of 2.00 x 10^-6 mol/L AgNO3 will result in the formation of a precipitate of AgCl.

To determine if a precipitate will form when you mix two solutions, you need to compare the ion concentrations in the resulting mixture with the solubility product constant (Ksp) of the potential precipitate.

1. Calculate the moles of NaCl and AgNO3:
Moles of NaCl = volume (in L) × concentration = 20.0 mL × 1.00 x 10^(-4) mol/L = 2.00 x 10^(-6) mol
Moles of AgNO3 = volume (in L) × concentration = 20.0 mL × 2.00 x 10^(-6) mol/L = 4.00 x 10^(-8) mol

2. Since both NaCl and AgNO3 fully dissociate into ions, the concentration of chloride ions (Cl-) in the final solution is twice the concentration of sodium chloride:
Concentration of Cl- = 2 × (2.00 x 10^(-6) mol/L) = 4.00 x 10^(-6) mol/L

3. Compare the concentration of chloride ions with the solubility product constant (Ksp) of AgCl. If the concentration of chloride ions is greater than the Ksp of AgCl, a precipitate will form.
[Cl-] > Ksp of AgCl
4.00 x 10^(-6) mol/L > 1.80 x 10^(-10)

Since the concentration of chloride ions (4.00 x 10^(-6) mol/L) is greater than the Ksp of AgCl (1.80 x 10^(-10)), a precipitate of AgCl will form.

You can conclude that a precipitate of AgCl will form when you mix 20.0 mL of 1.00 x 10^(-4) mol/L NaCl with 20.0 mL of 2.00 x 10^(-6) mol/L AgNO3.