1
Question
In 1 L saturated solution of AgCl[KSP(AgCl)=1.6×10–10],0.1 mol of CuCl[KSP(CuCl)=1.0×10–6] is added. The resultant concentration of Ag+ in the solution is 1.6×10–x. The value of "x" is
Open in App
Solution
AgCl→Ag+x+Cl−x+y
Ksp1=1.6×10−10=x(x+y) ....(i)
CuCl→Cu+y+Cl−y+x
Ksp2=1×10−6=y(x+y) ....(ii)
From equation (i) and (ii),
Ksp1Ksp2=1.6×10−4=xy ....(iii)
⇒x=1.6×10−4y
⇒Ksp1=1.6×10−10=1.6×10−4y(1.6×10−4y+y)
⇒10−6=y2(1.6×10−4+1)
⇒y=10−3⇒x=1.6×10−7
⇒[Ag+]=x=1.6×10−7
Ksp1=1.6×10−10=x(x+y) ....(i)
CuCl→Cu+y+Cl−y+x
Ksp2=1×10−6=y(x+y) ....(ii)
From equation (i) and (ii),
Ksp1Ksp2=1.6×10−4=xy ....(iii)
⇒x=1.6×10−4y
⇒Ksp1=1.6×10−10=1.6×10−4y(1.6×10−4y+y)
⇒10−6=y2(1.6×10−4+1)
⇒y=10−3⇒x=1.6×10−7
⇒[Ag+]=x=1.6×10−7
Suggest Corrections
1
Similar questions
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
