Question

Given ksp(agi) = 8.5 * 10-17.The solubility of agi in 0.1 m ki solution is

Answer 1

Hi,

Here is your answer

Dissociation of AgI is given by AgI----->Ag + I

Solubility of silver ions and Iodide ions is √8.5 x 10^-17

= 9.2 x 10^-9 moles

So by adding 0.1 KI, addition of iodide ions

new ksp is (9.2 x 10 ^-9) x 0.1

ksp = 9.2 x 10^-10

so new solubility is

√9.2 x 10^-10

= 3.0 x 10^-5 moles.

=