Chemistry

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(a) The solubility of bismuth iodide (BiI3) in water is 4.891 x 10-3 M. Calculate the value of the solubility product. The dissolution products are Bi^3+ (aq) and I^- (aq).


(b) Do you expect the solubility of (BiI3) in 0.1 M NaCL (aq) to be equal to, greater than, or less than that of BiI3 in water? Explain.

i) Greater than, the additional atomic species in solution allows for a greater entropy change upon mixing. Additionally, ionic bonds in solution are able to be formed between Bi/Cl and Na/I ions which would not be formed otherwise.

ii)Less than, the presence of ions already in solution results in the common ion effect, and thus the ability for new salts to be dissolved in solution is greatly diminished.

iii)Equal to, as even though there are already a number of cations and anions in solution they are of different chemical identify, and therefore will have no effect on solubility to first order.

  • Chemistry -

    b)III

  • Chemistry -

    (a) 10^-12*27*(x)^4

    replace x with the solubility you have

  • Chemistry -

    something is wrong. i couldn't find right answer

  • Chemistry -

    (a) 10^-12*27*(x)^4

    replace x with the solubility you have

    10^-12*27(4.891 x 10-3)^4
    x=4.891x10^-3 (as taken from a)*is this the solubility

    1.54509225456e-20 (it was wrong)

    What when wrong?

  • Chemistry -

    you get wrong, qpt

  • Chemistry -

    Yup it gave me a big red X.

  • Chemistry -

    Try without 10^-12

  • Chemistry -

    (x) * ((x*3))^3

  • Chemistry -

    3.27*10^-7

  • Chemistry -

    27*(4.891*10^-3)^4

  • Chemistry -

    thank you

  • Chemistry -

    1.54e-8

  • Chemistry -

    5.043*10^-7

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