Answer :
Answer:
[tex][Ag^+]=2.82x10^{-4}M[/tex]
Explanation:
Hello there!
In this case, for the ionization of silver iodide we have:
[tex]AgI(s)\rightleftharpoons Ag^+(aq)+I^-(aq)\\\\Ksp=[Ag^+][I^-][/tex]
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:
[tex][I^-]=[H^+]=10^{-3.55}=2.82x10^{-4}M[/tex]
Now, we can set up the equilibrium expression as shown below:
[tex]Ksp=8.51x10^{-17}=(x)(x+2.82x10^{-4})[/tex]
Thus, by solving for x which stands for the concentration of both silver and iodide ions at equilibrium, we have:
[tex]x=[Ag^+]=2.82x10^{-4}M[/tex]
Best regards!