Addition of 12M (concentrated) HCl to saturated NaCl results in precipitation of NaCl.?

Addition of 12M (concentrated) HCl to saturated NaCl results in precipitation of NaCl. Yet when 1M HCl is added to the same NaCl solution, no precipitate forms. Please Explain.

Thanks

Answer:
The solubility of NaCl in water is about 6M. So the solubility product constant will be:

Ksp=[Na+][Cl-]=6M*6M=36M^2

When you add concentrated HCl, what you are really adding is H+ + Cl-, and you are pushing the [Cl-] higher. So, to keep Ksp the same, Na+ has to come out of solution in the form of NaCl. If you know the exact volumes involved, you will be able to calculate the amount of NaCl that will come out of solution.

When you add dilute HCl, you are still adding Cl- as well, but you are also adding a lot more water, thus lowering the concentration of all the ions in the solution. So, [Na+][Cl-] will be less than Ksp and nothing will precipitiate.
This is a classic example of the Common Ion Effect, which you should look up.

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