Question

Compare the solubility, in milligrams per milliliter, of silver iodide, AgI, in water that is 0.020M in AgNO3

*******answer : 1.0x10^-12   

but how do i arrive at this answer? can you show me the work?

Answer 1

We first need:

solubility of AgI in water

then

solubility of AgI in 0.02 M of AgNO3

let us first start with water:

a)

solubility of wateR:

AgI(s) <-> Ag+ +I.

Ksp = [Ag+][I-]

assume concentrations are:

S = AgI = Ag+ = I-

so

Ksp = S*S

S = sqrt(Ksp)

Ksp of AgI = 1.5*10^-16

so

S = sqrt(1.5*10^-16) =1.224*10^-8 mol of AgI / Liter

in pure water....

now...

for

b)

AgNO3 = 0.02 M solution:

AgI(s) <-> Ag+ +I.

Ksp = [Ag+][I-]

in this case

[Ag+] = 0.02 M is set

so

Ksp = [AG+][I-]

1.5*10^-16 = (0.02)(S)

Assume 1 mol of I- is solubility

swo

[S] = (1.5*10^-16) / 0.02 = 7.5*10^-15

so

Comparison:

Ratio = S of AgI in AgNO3 solution / S of AgI in water = (7.5*10^-15) / (1.224*10^-8) = 6.1274*10^-7

Note that it is 6.1274*10^-7 less soluble in AgNO3 solution than in pure water

NOTE: please provide specific Ksp value for your data, Ksp data is pretty sensible, specifically with respect to AgI