The ksp of Ba(IO3)2 at 25 is 6×10^10 . What is the molar solubility of Ba(IO3)2 ?

4 Answers

  • The balanced equation is:

    Ba(IO3)2 <-> Ba2 + 2IO3

    I….._………….0……..0

    C…_………..+x……+2x

    F…_………….x………2x

    Once balanced, input information at equilibrium:

    Ksp = [Ba][IO3]^2

    6.0×10^-10 = (x)(2x)^2

    6.0×10^-10 = 4x^3

    x = 0.00053 M = 0.00053 mol/L

    This value for x also equals the molar solubility (mol/L) of Ba(IO3)2.

  • Ksp = [Ba+2][IO3]^2

    If x g-ions of Ba+2 are in solution, so are 2x g-ions of the iodate. Then

    4x^3= 6×10^-10 or x^3=150x^-12. Then x=5.2×10^-4, which is the molar solubility of the salt.

  • Ba Io3 2

  • For the best answers, search on this site https://shorturl.im/k8eHP

    Ksp = [Ba^2+(aq)][CrO4^2-(aq)] = 2.1*10^-10 [CrO4^2-(aq)] = 2.1*10^-10 / [Ba^2+(aq)] Since BaCrO4 has very low solubility, we can assume that in 0.0015M Na2CrO4(aq), [CrO4^2-(aq)] = 0.0015 mol/litre. So, 1.5*10^-3 = 2.1*10^-10 /[Ba^2+(aq)] then, [Ba^2+(aq)] = 2.1*10^-10 / 1.5*10^-3 = 1.4*10^-7 mol/litre. On dissolving BaCrO4, 1 mole of BaCrO4 produces 1 mole of Ba^2+ ions Therefore, Solubility of BaCrO4 = 1.4*10^-7 mol/litre Since BaCrO4 has Mr = 253 Solubility = 253 * 1.4*10^-7 g/litre = 3.54*10^-5 g/litre.

Leave a Comment