Question

How many moles of BaF2 are soluble in 250ml of sodium fluoride (NaF)? Concentration of NaF = 0.12 mol/

pKsp of BaF2 = 5.8

Answer 1

0.0000278 moles of barium fluoride are soluble in 250ml of sodium fluoride.

Step-by-step explanation:

Concentration of sodium fluoride (NaF) = 0.12 mol/L

`NaF\rightarrow Na^++F^-`

1 mole of NaF gives 1 mole of fluoride ions.

Concentration of fluoride ions from NaF=
`[F^-]=0.12 mol/L`

Concentration of barium fluoride:

`p_{K_(sp)}` of barium fluoride = 5.8

`p_{K_(sp)}=-\log[K_(sp)]`

`5.8=-\log[K_(sp)]`

Solubility product of barium fluoride :
`K_(sp)`

`K_(sp)=1.5848* 10^(-6)`

`BaF_2\rightleftharpoons Ba^(2+)+2F^-`

S (2S+ 0.12)

Total Concentration of barium ions in a final solution = S

Total Concentration of fluoride ions in a final solution= (2S+ 0.12)

An expression of
`K_(sp)` is given as:

`K_(sp)=S* (2S+0.12)^2`

`1.5848* 10^(-6)=4S^3+0.0144S+0.48S^2`

Since, barium fluoride is sparingly soluble salt. Value of S will be small so we can neglect
`S^3 \& S^2` values

`1.5848* 10^(-6)=0.0144S`

Calculating for S;

S = 0.00011 mol/L

`Ba^(2+)=S=0.00011 mol/L`

According to reaction , 1 mol of barium fluoride gives 1 mole of barium ion.Then 0.00011 barium ions will obtained from 0.00011 mol/L of barium fluoride solution.

So the concentration of barium fluoride is 0.00011 mol/L.

Volume of the solution = 250 ml = 0.250 L

`0.00011 mol/L=(n)/(0.250 L)`

n = 0.0000278 mol

0.0000278 moles of barium fluoride are soluble in 250ml of sodium fluoride.