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

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.

Step-by-step 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* 10^(-6)`

`K_(sp)=[Ba^(2+)][F^-]^2`

`1.6* 10^(-6)=[0.016 M]* [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=(0.01 M* 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.