An aqueous solution of sodium fluoride is slowly added to a water sample that contains barium ion (2.75×10-2m ) and calcium ion (6.70×10-2m ). what is the remaining concentration of the first ion to precipitate when the second ion begins to precipitate?

Answer : The remaining concentration of the first ion to precipitate when the second ion begins to precipitate is,

Explanation :

The dissociation of barium fluoride is written as:

The expression for solubility constant for this reaction will be,

As we know that at room temperature the of barium fluoride is, .

Now put all the given values in this expression, we get:

The barium fluoride precipitate when fluoride ion is equal to 0.00603 M.

The dissociation of calcium fluoride is written as:

The expression for solubility constant for this reaction will be,

As we know that at room temperature the of calcium fluoride is, .

Now put all the given values in this expression, we get:

The calcium fluoride precipitate when fluoride ion is equal to 0.0000241 M.

Since, the fluoride ion concentration in calcium fluoride is less then the fluoride ion concentration in barium fluoride. That means, calcium fluoride will precipitate first.

Thus, the concentration ion remaining at concentration (0.00603 M) is calculated as:

Now put all the given values in this expression, we get:

Therefore, the remaining concentration of the first ion to precipitate when the second ion begins to precipitate is,

i think   the answer is c hope i'm right