What is the molar solubility of CaF2 in 0.0022 M Ca(NO3)2? (Ksp = 3.9 x 10-11) CaF2 (s) D Ca2+ (aq) + 2 F- (aq) ?

(Hint: contain common ion Ca2+, concentration of Ca2+ will not be zero at the initial step)

(The answer should be in 3 significant numbers such as 1.50E-03 or 0.00150 for 1.50 x 10-3) 

1 Answer

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  • Bobby
    Lv 7
    2 months ago

    CaF2 = Ca2+ + 2F-

    Ksp = [Ca2+] *[F-] ^2 = 3.9 x 10-11

    if x mole of CaF2 dissolves 

    ICE table .......[Ca2+] [F-]

    I .......................0.0022 ...0

    C.....................+x.........+2x

    E.................0.0022 +x ...2x

    substituting in Ksp we have 

    3.9 x 10-11= (0.0022+x) * 4x^2 

    We assume that x << 0.0022, so that 0.0022 + x = 0.0022

    0.0088x^2 = 3.9 x 10-11

    x^2 = 3.9 x 10-11 / 0.0088 = 4.43182E-09

    x= 6.66 * 10 ^-5

    molar solubility of CaF2 in 0.0022 M Ca2+ = 6.66 * 10 ^-5 moles/ liter 

    there is your answer 

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