# Calculate the molar solubility of barium fluoride in each of the following.?

Calculate the molar solubility of barium fluoride in each of the following.?

• A) in pure water :

BaF2 (s) <——-> Ba(2+) (aq) + 2F(-) (aq)

let the solubility of BaF2 be s moles/L …the the solution will contain s moles of Ba(2+) and 2s moles of F(-) ions respectively per litre …hence the solubility product Ksp of BaF2 would be given by the expression :

Ksp = [Ba2+] [F-]^2

Ksp = s(2s)^2 = s X 4s^2 = 4s^3

but the given value of Ksp is 2.45 X 10^-5

so 4s^3 = 2.45 X 10^-5

s^3 = 2.45 X 10^-5/4 = 6.125 X 10^-6

s = 1.83 X 10^-2 mole/L

B)in 0.12 M Ba(NO3)2…

Ba(NO3)2 dissociates completely ,therefore in 0.12 M Ba(NO3)2 solution …[Ba2+] = 0.12 M

so now concentration of Ba(2+) will be (s+ 0.12) moles/litre or 0.12 moles/litre approximately ..

so again using the same expression…

Ksp = [Ba2+] [F-]^2

putting the values..

2.45 X 10^-5 = 0.12 X (2s)^2

2.45 X 10^-5 = 0.12 X 4s^2

2.45 X 10^-5 = 0.48 X s^2

5.104 X 10^-5 = s^2

s = 7.144 X 10^-3 mole/L

C) similarly in 0.15 M NaF …concetration of F(-) will be (0.15 + 2s) moles/L or 0.15 moles/L approximately..

so as Ksp = [Ba2+] [F-]^2

2.45 X 10^-5 = s X (0.15)^2

2.45 X 10^-5 = s X 0.0225

s = 2.45 X 10^-5/0.0225 = 1.09 X 10^-3 moles/L

hey …did you remembered me ?? i gave that answer to your SO2Cl2 partial pressure problem and unfortunately it was wrong …

here is that question —

but anyways i have corrected my answer you can see that question just see those comments…….

i forgot to change the units of pressure in atm…

i am extremely sorry for my wrong answer ….!!!

• Solubility Of Barium

• BaF2 –> Ba + 2F Ksp = 1.7 x 10^-6 Ksp = [Ba][F]^2 1.) 7.52 x 10^-3 mol/L 2.) 2.04 x 10^-3 mol/L 3.) 1.06 x 10^-5 mol/L assumption: MNaF ionizes completely in reaction. I feel that it may have been a typo as there is no element: “M”

• solubility of a substance means no. of gm. of that substance present in 100 ml of solution.

Molar solubility means no. of moles in 100ml of sol.

0.12M Ba(NO3)2 = 0.012 moles /100ml , so molar solubility of Ba(NO3)2 is 0.012moles.

IIly for 0.15 M NaF = 0.015 moles /100ml i.e. molar solubility is 0.015 moles.