Question

What is the concentration of silver ions where silver iodide, Agl, is in a solution of hydroiodic

acid, HI, that has a pH of 3.55? Ksp = 8.51x10-17

Answer 1

`[Ag^+]=2.82x10^(-4)M`

Step-by-step explanation:

Hello there!

In this case, for the ionization of silver iodide we have:

`AgI(s)\rightleftharpoons Ag^+(aq)+I^-(aq)\\\\Ksp=[Ag^+][I^-]`

Now, since we have the effect of iodide ions from the HI, it is possible to compute that concentration as that of the hydrogen ions equals that of the iodide ones:

`[I^-]=[H^+]=10^(-3.55)=2.82x10^(-4)M`

Now, we can set up the equilibrium expression as shown below:

`Ksp=8.51x10^(-17)=(x)(x+2.82x10^(-4))`

Thus, by solving for x which stands for the concentration of both silver and iodide ions at equilibrium, we have:

`x=[Ag^+]=2.82x10^(-4)M`

Best regards!