Question

AgCl (s) + --> <-- Ag + (aq) + Cl- (aq) Shown above is information about the dissolution of AgCl in water at 298 K. In a chemistry lab a student wants to determine the value of s, the molar solubility of AgCl, by measuring [Ag+] in a saturated solution prepared by mixing excess AgCl and distilled water. How would the results of the experiment be altered if the student mixed excess AgCl with tap water (in which [Cl-] = .01 M) instead of distilled water and the student did not account for the Cl- in the tap water?

A. The value obtained for Ksp would be too small because Cl- (aq) ions would be attracted to the Ag+ ions in the AgCl crystals, thus preventing water molecules from reaching the crystals.

B The value obtained for Ksp would be too small because less AgCl (s) would dissolve because of the common ion effect due to the Cl- already in the water.

C The value obtained for Ksp would be too large because AgCl (s) more would dissolve because of the attractions between Ag+ ions in the AgCl crystals and the Cl- ions in the water.

D The results of the experiment would not be altered because .01 M is such a small concentration of Cl- i(aq) ons and thus has no effect on the dissolution of AgCl (s).

Answer 1

Option B is correct , The value obtained for Ksp would be too small because less AgCl (s) would dissolve because of the common ion effect due to the Cl- already in the water.

Here I have given the reasons in which results affected is explained

Reason : 1

[ Cl- ] = Concentration of Chloride ion

[Ag+ ] = Concentration of silver ion

K = [Ag+ ] [Cl- ] / [AgCl]

but there is no conc for pure solid [AgCl] so just write

Ksp = [Ag+ ] [ Cl- ]

Ksp (AgCl) = 1.7 x10-10 ( from appendix J , list of Ksp for ionic compounds)

Let solubility at equilibrium = ‘S’ = [ Cl- ] = [Ag+ ]

Ksp(AgCl) = S2 = 1.7 x10-10

There fore, S = square root of 1.7 x10-10 = [ Cl- ] = [Ag+ ] = 0.0000130

[Ag+]= 0.0000130 = 1.3 x 10-5

To dissolve Silver chloride in 0.0100 M chlorine containing tap water instead of in pure distilled water, what is the equilibrium concentration of the Silver is change this time.

The calculations are different from before. This time the concentration of the chloride ions is generated by the Cl- in tap water. This time the calculation is changed as follows :

So we assume:

[Cl−]=0.0100M in Tap water

The rest calculation be like this:

Ksp (AgCl) = 1.7 x10-10 ( from appendix J , list of Ksp for ionic compounds)

Let solubility at equilibrium = ‘S’ = [Ag+ ]

Ksp(AgCl) = Sx 0.01x [Cl−] = 1.7 x10-10

[Ag+] = S = 1.7 x10-10 /0.01 = 1.3 x10-13

therefore:

Finally, compare that value for AgCl solubility :

Solution in 0.0100 M [Cl-] tap water:

Now, [Ag+] = S = 1.7 x10-10 /0.01*1.3 x 10-5 = 1.3 x10-13 M

Original solution:

Then [Ag+]= 0.0000130 = 1.3 x 10-5 M

The concentration of the Silver ions has Decreased by a factor of about 108. If more concentrated solutions of chlorides present in tap water is used, the solubility decreases further.

Reason : 2

Effect of : Common Ion Effect on Solubility

Addition of common ion effects solubility & decreases it , as the reaction shifts toward the left to relieve excess product stress from the reaction. Addtion of a common ion in reaction causes the equilibrium to shift left, toward the reactants, causing precipitation.