Respuesta :
Answer:
0.0000278 moles of barium fluoride are soluble in 250ml of sodium fluoride.
Explanation:
Concentration of sodium fluoride (NaF) = 0.12 mol/L
[tex]NaF\rightarrow Na^++F^-[/tex]
1 mole of NaF gives 1 mole of fluoride ions.
Concentration of fluoride ions from NaF= [tex][F^-]=0.12 mol/L[/tex]
Concentration of barium fluoride:
[tex]p_{K_{sp}}[/tex] of barium fluoride = 5.8
[tex]p_{K_{sp}}=-\log[K_{sp}][/tex]
[tex]5.8=-\log[K_{sp}][/tex]
Solubility product of barium fluoride :[tex]K_{sp}[/tex]
[tex]K_{sp}=1.5848\times 10^{-6}[/tex]
[tex]BaF_2\rightleftharpoons Ba^{2+}+2F^-[/tex]
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 [tex]K_{sp}[/tex] is given as:
[tex]K_{sp}=S\times (2S+0.12)^2[/tex]
[tex]1.5848\times 10^{-6}=4S^3+0.0144S+0.48S^2[/tex]
Since, barium fluoride is sparingly soluble salt. Value of S will be small so we can neglect [tex]S^3 \& S^2[/tex] values
[tex]1.5848\times 10^{-6}=0.0144S[/tex]
Calculating for S;
S = 0.00011 mol/L
[tex]Ba^{2+}=S=0.00011 mol/L[/tex]
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
[tex]0.00011 mol/L=\frac{n}{0.250 L}[/tex]
n = 0.0000278 mol
0.0000278 moles of barium fluoride are soluble in 250ml of sodium fluoride.
