Tough questions series – 2006 CHEM 13 NEWS Exam

(This is a reprint from the May 2007 issue of Chem 13 News, page 16.)

How many moles of solid sodium fluoride should be added to 1.0 L of a saturated solution of barium fluoride, BaF2, at 25oC to raise the fluoride concentration to 3.0×10−2 mol L−1?  The solubility product constant for BaF2 is Ksp = 1.7×10−6 at 25oC.

On the 2006 CHEM 13 NEWS Exam, the above question stumped practically all students.  Only 11% of the students answered this question correctly, and about 60% entered no answer at all.  For most of the other questions, the percentage of students who entered “no answer” ranged from 3% to 30%.  

The CHEM 13 NEWS Exam consists of 40 multiple-choice questions intended for a well-read senior high school chemistry student.  Carey Bissonnette, one of the exam authors, stated “We knew that students would find this question difficult.  To obtain the answer, one must take account of the fact that adding NaF to a saturated solution of BaF2 will cause BaF2 to precipitate.”  The following are two approaches to solving this problem.  

Approach #1:  In a saturated solution of BaF2, [F] = 2 [Ba2+] and Ksp = [Ba2+] [F]2 , so

The concentration of Ba2+ equals the cubed root of the solubility constant, Ksp, divided by 4, which equals 0.007518 moles per litre and the concentration of F- equals 0.1504 moles per litre.

When NaF is added, BaF2 will precipitate from the solution.  Let “x” be the number of moles of NaF added and let “z” be the amount of BaF2 that precipitates.

Starting with the reaction and then adding in the calculated concentrations above, the steps to solving for x and z are as follows:

Ba2+(aq)    +    2 F(aq)    →    BaF2(s)
Reactants/Products Ba2+(aq) 2 F(aq) BaF2(s)
Initial (mol) 0.007518 0.01504 + x -
Final (mol) 0.007518 − z 0.01504 + x − 2z -

But the final concentration of F is 3.0×10−2 mol L−1.  Therefore, the final concentration of Ba2+ must be:

[Ba2+]=Ksp/(3.0x10-2)2 = 0.001889 mol-1.

 So, we have:

0.007518 − z  =  0.001889                           ⇒    z = 0.005629


0.01504 + x − 2(0.005629)  =  3.0×10−2     ⇒    x = 0.026 mol L−1

Therefore, we must add 0.026 moles of NaF.

Approach #2:  Use charge balance! (Not normally taught in high school, but a simple and useful concept.)  The positive charge in solution must offset the negative charge.  

In equation form:  [Na+] + 2 [Ba2+] = [F]. (Note: There is a “2” in front of [Ba2+] because 1 mol of Ba2+ carries twice as much positive charge as does 1 mol of Na+.)  

Since Ksp = [Ba2+] [F]2  and  [F] = 3.0×10−2 mol L−1, we can write:

       [Na+]   +   2 Ksp/(3.0×10−2)2  =  3.0x10−2
       [Na+]  =  0.026 mol L−1

But [Na+] tells us exactly how much NaF must be added to the solution.