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

SOMEONE ASKED 👇

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

HERE THE ANSWERS 👇

  • 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 —

    http://answers.yahoo.com/question/index;_ylt=AiWVd…

    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.

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