Question

The Ksp of barium fluoride, BaF2, is 1.6 x 10-6. A solution of 5.0M NaF is added dropwise to a 2.0L solution that is 0.016 M in Ba2+. When the concentration of fluoride ion exceeds M, BaF2 will precipitate. What volume (in mL) of NaF must be added to cause BaF2 to precipitate? mL

Answer 1

Answer:

Precipitation will occur after the concentration of fluoride ions exceeds 0.01 M.

4 mL volume of NaF must be added to cause $BaF_2$ to precipitate.

Explanation:

Concentration of barium ions = $[Ba^{2+}]=0.016 M$

Volume of barium ion solution = 2.0 L

$BaF_2\rightleftharpoons Ba^{2+}+2F^-$

The solubility product of the barium fluoride = $K_{sp}=1.6\times 10^{-6}$

$K_{sp}=[Ba^{2+}][F^-]^2$

$1.6\times 10^{-6}=[0.016 M]\times [F^-]^2$

$[F^-]=0.01 M$

Precipitation will occur after the concentration of fluoride ions exceeds 0.01 M.

Concentration of fluoride ion = $C_1=0.01 M$

Volume of solution = $V_1=2.0 L$

Concentration of NaF solution added = $C_2=5.0 M$

Volume of NaF solution added = $V_2=?$

$C_1V_1=C_2V_2$

$V_2=\frac{0.01 M\times 2.0 L}{5.0 M}=0.004L$

1 L = 1000 mL

0.004 L = 0.004 × 1000 mL = 4 mL

4 mL volume of NaF must be added to cause $BaF_2$ to precipitate.