Question

Part 1: The Ksp for silver chloride (AgCl) is 1.77 × 10−10. Which of the following represents the Ksp expression for this substance?

A Ksp = [AgCl]/[Ag+][Cl-]

B Ksp = 1/[Ag+][Cl-]

C Ksp = [Ag+][Cl-]

D Ksp = [Ag+][Cl-]/[AgCl]

The answer was C

Part 2: You add 72.0 mL of a 0.472 M AgNO3 solution to 127.1 mL of a 0.114 M NaCl solution. Determine the concentration of Ag+ and Cl- ions in solution as if all remain dissolved. After, use your Ksp expression to determine if a precipitate will form. Which of the following represents what will occur and why?

A- A precipitation will form, because Q>K

B- A precipitation will not form, because Q < K

C- A precipitation will form, because Q < K

D- A precipitation will not form, because Q>K

This answer was A.

Part 3: Create an ICE table for the FORMATION of AgCl. React the two ions COMPLETELY, and then determine which ion is the limiting reagent. Finally, determine the concentration (in molarity) of the ion remaining, and enter the value below.

Part 4: Finally, create another ICE table for the dissolution of AgCl. Use the concentration from part 3 for the INITIAL concentration of ion remaining in part 3 and 0.000 M for the INITIAL concentration of the other ion. Finally, calculate the equilibrium concentration of both ions. Enter the equilibrium concentration of the ion that was the limiting reagent in part 3 below.

I was able to figure out the first two questions but can't seem to figure out the last two.

Answer 1

Part A

AgCl ==== Ag+ + Cl-

Ksp is the equilibrium constant of solid substance dissolved in aqueous solution, so AgCl will formed Ag+ and Cl- with one coefficient each, hence the Ksp will be simply the product of Ag+ and Cl- concentration

Ksp = [Ag+][Cl-]

Part B

Reaction will be

AgNO3 + NaCl ----- AgCl + NaNO3

Number of moles of AgNO3 = volume of solution (in L) * Molarity = 72/1000 * 0.472 = 0.033984 moles

Number of moles of NaCl = volume of solution (in L) * Molarity = 127.1/1000 * 0.114 = 0.0144894 moles

Concentration of Ag+ = 0.033984/(72+127.1) * 1000 = 0.17068M

Concentration of Cl- = 0.0144894/(72+127.1) * 1000 = 0.07277448518M

Q= (0.17068)(0.0727748518) = 1.242 * 10^(-2)

Since Q > Ksp, hence the precipirate will form (Ksp = 1.77 * 10^(-10) for AgCl)

Part 3

Limting reagent = Cl-

moles of Ag+ left = 0.033984 - 0.0144894 = 0.0194946

Molarity of Ag+ left = 0.0194946/199.1 * 1000 = 0.0979M

Part 4

AgCl ------ Ag+ + Cl-

Initial concentration of Ag+ = 0.0979M

Initial concentration of Cl- = 0M

Data is missing there should be some K value given in the problem