Find the molar solubility of PbI2, Ksp = 7.9 x 10–9 in each of the following: (a) Pure water (b) 0.10 M NaI (c) 0.10 M Pb(NO3)2?

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  • Bobby
    Lv 7
    2 months ago

    Ksp = [Pb2+]*[I-]^2

     7.9 x 10–9 =  [Pb2+] * [I-]^2

    a) if x mole of PbI2 dissolves in pure water we have 

    7.9 x 10–9 = x* 4x^2

    x^3 = 7.9 x 10–9 / 4

    molar solubility of PbI2 = 1.25 * 10 ^-3 moles / liter  

    b) if the water contains 0.1 M I- we have 

    7.9 x 10–9 = x * (2x+0.1)^2

    we know 2x is very small so we ignore it and get 

    7.9 x 10–9 = x * 0.1^2

    molar solubility of PbI2 = 7.9 * 10 ^-7 Moles / liter 

    c) if the water contains 0.1 M Pb2+ we have

     7.9 x 10–9 = (x + .1) * (2x)^2

    we know x is very small so we have 

    7.9 x 10–9 = 0.4 x^2

    x^2 =  7.9 x 10–9 /4 = 1.975E-09

    molar solubility of PbI2 = 4.44 * 10 ^-5 Moles / liter 

    hcbiochem gave 7.9X10^-7 = (0.10)(2x)^2 

    this is a mistake should be 10*-9 

    it is more respectful to point out errors than nasty thumbs down 

  • 2 months ago

    PbI2(s) <--> Pb2+(aq) + 2 I-(aq)

    Ksp = 7.9X10^-9 = [Pb2+][I-]^2

    (a) Let x = [Pb2+] = molar solubility, and 2x = [I-]. Then,

    7.9X10^-9 = x(2x)^2 = 4x^3

    x = molar solubility = 1.25X10^-3 M

    (b) Let x = [Pb2+] = molar solubility, and 0.10 + 2x = [I-]. Because Ksp is small, assume that 2x will be small compared to 0.10. Then,

    7.9X10^-9 = x(0.10)^2

    x = 7.9X10^-7 M

    (c) Let x = molar solubility. Then, 2x = [I-] and 0.1 + x = [Pb2+]. Assume x will be small compared to 0.10. Then,

    7.9X10^-7 = (0.10)(2x)^2

    x = molar solubility = 1.4X10^-4 M

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