Equal volumes of 0.240 M AgNO3 and 0.140 M ZnCl2 solution are mixed. Calculate the quilibrium concentrations of Ag+, and Zn2+.

ZnCl2(aq) + 2AgNO3(aq) ==> Zn(NO3)2(aq) + 2AgCl(s)

I would assume a convenient volume of each, say 100 mL, then
(ZnCl2) = (Zn^+2) = 100mL x 0.140M = 14 mmoles/200 mL = 0.070M Zn ion.
(Cl^-) = 2 x 14 = 28 mmoles Cl^-

The Ag^+ is trickier because AgCl is insoluble AND you have some of the Cl^-, which is in excess, acting as a common ion which decreases the solubility of AgCl.
AgCl(s) ==> Ag^+ + Cl^-
Ksp = (Ag^+)(Cl^-) = 1.82E-10
(Ag^+) = Ksp/(Cl^-).
You know Ksp. (Cl^-) is done this way. AgCl formed is 24 mmoles which uses up all of the Ag in the AgCl and 24 of the 28 Cl^- in the ZnCl2. So there are 4 mmoles Cl^- in excess and (Cl^-) = mmoles/mL = 4 mmoles/200 mL = ??
Then substitute that into the (Ag^+) = Ksp/(Cl^-) and solve for Ag^+. I don't know if this was a trick question your prof gave you or if your prof wanted you to be on your toes and recognize that a ppt was formed and there was a common ion present.

To calculate the equilibrium concentrations of Ag+ and Zn2+ ions, we need to use the concept of stoichiometry and the reaction equation between AgNO3 and ZnCl2.

The reaction equation between AgNO3 and ZnCl2 is:

AgNO3 + ZnCl2 → AgCl + Zn(NO3)2

Based on the balanced equation, we can see that one mole of AgNO3 reacts with one mole of ZnCl2 to form one mole of AgCl and one mole of Zn(NO3)2.

Given the information that we have equal volumes of 0.240 M AgNO3 and 0.140 M ZnCl2 solutions being mixed, we can assume that the volumes of the two solutions are equal. Let's say the volume of each solution is V.

Now, we can calculate the moles of AgNO3 and ZnCl2 using the given concentrations and volumes.

Moles of AgNO3 in the mixture = Concentration (0.240 M) × Volume (V)
Moles of ZnCl2 in the mixture = Concentration (0.140 M) × Volume (V)

Since we have assumed equal volumes, the number of moles of AgNO3 and ZnCl2 will be the same.

Now, since one mole of AgNO3 reacts with one mole of ZnCl2, the number of moles of AgCl formed will also be the same.

The balanced equation tells us that one mole of AgCl produces one mole of Ag+ and one mole of Zn2+. Therefore, the equilibrium concentration of Ag+ and Zn2+ will also be equal to the number of moles.

Equilibrium concentration of Ag+ = Number of moles of Ag+
Equilibrium concentration of Zn2+ = Number of moles of Zn2+

Hence, the equilibrium concentrations of Ag+ and Zn2+ ions will be equal to the number of moles of AgNO3 and ZnCl2, respectively, which can be calculated by multiplying their respective concentrations with the assumption of equal volumes.