Question

Use the following information to calculate the solubility product constant, Ksp, for CuCl. A saturated solution of CuCl in water was prepared and filtered. From the filtrate, 1.0 L was measured out into a beaker and evaporated to dryness. The solid CuCl residue recovered in the beaker was found to weigh 0.041g.

A. Ksp =1.7 × 10¯9
B. Ksp = 1.7 × 10¯7
C. Ksp = 1.7 × 10¯5
D. Ksp = 4.3 × 10¯4
E. Ksp = 2.1 × 10¯2

Answer 1

B. Ksp = 1.7 × 10¯⁷

Step-by-step explanation:

Hello there!

In this case, for this solubility equilibrium problem, we first need to set up the chemical reaction describing the dissolution of the involved salt, CuCl:

`CuCl(s)\rightleftharpoons Cu^+(aq)+Cl^-(aq)`

Next, we write the corresponding equilibrium expression:

`Ksp=[Cu^+][Cl^-]`

Now, we need to calculate the concentrations of copper (I) and chloride ions at equilibrium; thus, given that 0.041 g of this solid is completely dissolved in 1.0 L of solution, we can firstly calculate the moles present in the solution:

`n_(CuCl)=0.041gCuCl*(1molCuCl)/(99gCuCl) =4.14x10^-4mol`

Afterwards, since all the species in the reaction, CuCl, Cu+ and Cl- are in a 1:1:1 mole ratio, we realize that those moles correspond to ions in the solution, so their concentrations are:

`[Cu^+]=[Cl^-]=(4.14x10^(-4)mol)/(1.0L)= 4.14x10^(-4)M`

Then, we compute the Ksp by plug this value in the equilibrium expression:

`Ksp=(4.14x10^(-4))(4.14x10^(-4))=1.7x10^(-7)`

Thus, the answer would be B. Ksp = 1.7 × 10¯⁷.

Regards!