Question

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

Answer 1

Answer:

$[Ag^+]=2.82x10^{-4}M$

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:

$x=[Ag^+]=2.82x10^{-4}M$

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