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# A solution has a Mg^2+ concentration of 0.010 mol/l. Will Mg(OH)2 precipitate if the OH^- concentration of the solution is (i) 10^-5 mol/l (ii) 10^- mol/L.

Aman Bansal
592 Points
8 years ago

Dear Nancy,

the source of the chloride is unimportant (at this level). Let us assume the chloride came from some dissolved sodium chloride, sufficient to make the solution 0.0100 M. So, on to the solution . . .

The dissociation equation for AgCl is:

AgCl (s) <===> Ag+ (aq) + Cl¯ (aq)

The Ksp expression is:

Ksp = [Ag+] [Cl¯]

This is the equation we must solve. First we put in the Ksp value:

1.77 x 10¯10 = [Ag+] [Cl¯]

Now, we have to reason out the values of the two guys on the right. The problem specifies that [Cl¯] is already 0.0100. I get another ''x'' amount from the dissolving AgCl. Of course, [Ag+] is ''x.''

Substituting, we get:

1.77 x 10¯10 = (x) (0.0100 + x)

This will wind up to be a quadratic equation which is solvable via the quadratic formula. However, there is a chemical way to solve this problem. We reason that ''x'' is a small number, such that ''0.0100 + x'' is almost exactly equal to 0.0100. If we were to use 0.0100 rather than ''0.0100 + x,'' we would get essentially the same answer and do so much faster. So the problem becomes:

1.77 x 10¯10 = (x) (0.0100)

and

x = 1.77 x 10¯8 M

There is another reason why neglecting the ''x'' in ''0.0100 + x'' is OK. It turns out that measuring Ksp values are fairly difficult to do and, hence, have a fair amount of error already built into the value. So the very slight difference between ''x'' and ''0.0100 + x'' really has no bearing on the accuracy of the final answer. Why not? Because the Ksp already has significant error in it to begin with. Our "adding" a bit more error is insignificant compared to the error already there.

Example #2: The Ksp for silver carbonate is 8.4 x 10¯12. The concentration of carbonate ions in a saturated solution is 1.28 x 10¯4 M. What is the concentration of silver ions?

Solution:

Dissociation equation:

Ag2CO3 (s) <===> 2 Ag+ (aq) + CO32¯ (aq)

Ksp expression:

Ksp = [Ag+]2 [CO32¯]

Let us substitue into the Ksp expression:

8.4 x 10¯12 = (x)2 (1.28 x 10¯4)

Note: I could have used (1.28 x 10¯4 + 0.5x) for the carbonate. See above discussion for why the 0.5x can be dropped.

Divide both sides by 1.28 x 10¯4 and then take the square root:

[Ag+] = x = 2.56 x 10¯4 M

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Thanks

Aman Bansal