TBR Chemistry 3.72

[StretchDoe]

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The question states

Enough CaCl2 is added to water so that not all of it dissolves, and thus some CaCl2 remains as a solid on the bottom of the flask. The [CI"] is then measured. Addition of which of the following to the solution will increase the chloride ion concentration ([CI"])?

A. Water
B. Silver nitrate
C. Calcium fluoride
D. Sodium phosphate

Answer is D. I got it right. Simply used the solubility rules that I know. Also there was a table one could use to figure it out. However I was curious, why couldn't water be a correct answer? Wouldn't adding more water dissolve CaCl2?

Thanks!

 

[theonlytycrane]

The Ksp would be unchanged (it only depends on the ion concentrations).

 

[StretchDoe]

The Ksp would be unchanged (it only depends on the ion concentrations).

I don't quite follow? What does the Ksp have to do with this exactly? Isn't the Ksp unchanged when using either water or Sodium Phosphate?

I know adding Sodium Phosphate forms NaCl, which dissolves and increases the Cl- concentration. But doesn't adding more water allow CaCl2 to dissolve which increases Cl- concentration?

 

[theonlytycrane]

CaCl2 <-> Ca(2+) + 2Cl-

Na+ and (PO4)3- are also in solution. (PO4)3- combines with some Ca(2+) to form (Ca)3(PO4)2 as below. Note that NaCl is totally ionic. This results in more CaCl2 dissociation, resulting in higher Cl- concentration. The amount of ion and solid forms are related to each of their solubility products (Ksp). The Ksp for CaCl2, in equation form, is [Ca2+][Cl-]^2.

(Ca)3(PO4)2 <-> 3Ca(2+) + 2(PO4)3-

Water is just the solvent, though, so increasing its concentration doesn't change the Ksp's of either reaction.

 

[StretchDoe]

CaCl2 <-> Ca(2+) + 2Cl-

Na+ and (PO4)3- are also in solution. (PO4)3- combines with some Ca(2+) to form (Ca)3(PO4)2 as below. Note that NaCl is totally ionic. This results in more CaCl2 dissociation, resulting in higher Cl- concentration. The amount of ion and solid forms are related to each of their solubility products (Ksp). The Ksp for CaCl2, in equation form, is [Ca2+][Cl-]^2.

(Ca)3(PO4)2 <-> 3Ca(2+) + 2(PO4)3-

Water is just the solvent, though, so increasing its concentration doesn't change the Ksp's of either reaction.

I got this for the most part, but doesn't Ksp only change with a change in temperature not concentrations.

 

[theonlytycrane]

I got this for the most part, but doesn't Ksp only change with a change in temperature not concentrations.

Yeah, you're right. I should have said that water has no part in the Ksp calculation. In the Ksp for CaCl2, though, since [Ca2+] decreases, [Cl-] increases for the solubility product to stay constant.

 

[BerkReviewTeach]

The reason water is not correct comes back to the wording of the question. If you add water, then more of the salt at the bottom of the flask will dissolve into solution (resulting in more dissolved chloride anions), but the question asks for "concentration" and not grams, atoms, or moles. The solution is saturated, so adding water cannot make it more concentrated in dissolved salt, only equally concentrated. Let's say we added enough water to double the volume of the solution, then we'd get twice as much Cl- dissociated into the solution. Twice as many ions divided by twice the volume equals the same concentration.