Chemistry Question: Precipitates?

Question

Sodium chloride is purified for use as table salt by adding HCl to a saturated solution of NaCl (317 grams/Liter). When 25.5 mL of 7.85 M HCl is added to 0.1 Liter of saturated NaCl solution, how many grams of purified NaCl percipitate?

Answer 1

The molecular weight of NaCl is 58.5
so the molar solubility is 317/58.5 =5.42 mole/L

Ksp =[Na+][Cl-]= 5.42*5.42 = 29.38

When you mix the two solutions the concentration of the ions change.
[Na+]o= M2V2/(V1+V2)= 5.42*0.1/(0.0255+0.1) = 4.3187 =4.32 M
[Cl-]o= (M1V1+M2V2) / (V1+V2)= (7.85*0.0255+ 5.42*0.1) / (0.0255+0.1) =5.91 M

.. .. .. .. .. .. NaCl <=> Na+ +Cl-
Initial .. .. .. .. .. .. .. .. 4.32 .. 5.91
React .. .. .. .. .. .. .. .. .. x .. .. x
Produce .. .. x
At Equil. .. .. .. .. .. 4.32-x . 5.91-x

Ksp =(4.32-x)(5.91-x)=29.38 =>
x^2 +10.23-3.8588 =0 =>
x= 0.36424= 0.364 mole/L form (the other solution is <0 and thus rejected)

mass= mole*MW= M*V*MW= 0.364*(0.1+0.0255)*58.5 = 2.672 g NaCl precipitate