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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

Sagot :

Sebassandinidn

Answer:

B. Ksp = 1.7 × 10¯⁷

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:

[tex]CuCl(s)\rightleftharpoons Cu^+(aq)+Cl^-(aq)[/tex]

Next, we write the corresponding equilibrium expression:

[tex]Ksp=[Cu^+][Cl^-][/tex]

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:

[tex]n_{CuCl}=0.041gCuCl*\frac{1molCuCl}{99gCuCl} =4.14x10^-4mol[/tex]

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:

[tex][Cu^+]=[Cl^-]=\frac{4.14x10^{-4}mol}{1.0L}= 4.14x10^{-4}M[/tex]

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

[tex]Ksp=(4.14x10^{-4})(4.14x10^{-4})=1.7x10^{-7}[/tex]

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

Regards!

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