The question happens to be in an incorrect order but the correct question can be seen below;
What concentration of [tex]SO^{2-}_3[/tex] is in equilibrium with [tex]Ag_2SO_{3(S)}[/tex] and [tex]7.10*10^{-3}M[/tex] [tex]Ag^+[/tex]? (The [tex]K_{sp}[/tex] of
Answer:
[tex]2.96*10^{-10}M[/tex]
Explanation:
The concentration of [tex]SO^{2-}_3[/tex] can be determined by using the solubility concept.
Given ionic solid is [tex]Ag_2SO_{3(S)}[/tex] ;
The Equilibrium Equation for the ionic compound will be:
[tex]Ag_2SO_{3(S)}[/tex] ⇄[tex]2Ag_{(aq)}[/tex] + [tex]SO^{2-}_3_{(aq)}[/tex]
Now, the solubility product ([tex]K_{sp}[/tex]) of the ionic compound will be;
[tex]K_{sp}[/tex] [tex]= [Ag^+]^{2}[SO^{2-}_3][/tex]
Given that;
the concentration [tex]Ag^+[/tex] is [tex]7.10*10^{-3}M[/tex] ; &
solubility product of the given ionic solid is [tex]1.5*10^{-14}[/tex]
∴
[tex]K_{sp}[/tex] [tex]= [Ag^+]^{2}[SO^{2-}_3][/tex]
[tex]1.5*10^{-14}[/tex] [tex]= (7.10*10^{-3})^2[/tex] [tex][SO^{2-}_3][/tex]
[tex][SO^{2-}_3][/tex] = [tex]\frac{1.5*10^{-14}}{ (7.10*10^{-3})^2}[/tex]
= 2.97560008 × 10⁻¹⁰
≅ [tex]2.96*10^{-10}M[/tex]
Thus, the concentration of [tex][SO^{2-}_3][/tex] is [tex]2.96*10^{-10}M[/tex]
