Question

(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.

Answer 1

b)III

Answer 2

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

replace x with the solubility you have

Answer 3

something is wrong. i couldn't find right answer

Answer 4

(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?

Answer 5

you get wrong, qpt

Answer 6

Yup it gave me a big red X.

Answer 7

Try without 10^-12

Answer 8

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

Answer 9

3.27*10^-7

Answer 10

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

Answer 11

thank you

Answer 12

1.54e-8

Answer 13

5.043*10^-7