For the reaction
`Cu(s)+2Ag^(o+)(aq)rarr Cu^(2+)(aq)+2Ag(s)`
Fill in the blanks in the following table for the three solution at equilibrium.
`{:("Solution",[Cu^(2+)(aq)],[Ag^(o+)(aq)],K L^(-1)),(,mol L^(-1),mol L^(-1),mol L^(-1)),(1.,(a),1.0xx10^(-9),2.0xx10^(15)),(2.,2.0xx10^(-7),1.0xx10^(-11),(b)),(3.,2.0xx10^(-2),(c),2.0xx10^(15)):}`

Answer

A, B, C

Explanation

a. `K=([Cu^(2+)])/([Ag^(o+)]^(2))`
`2.0xx10^(15)=([Cu^(2+)])/((1.0xx10^(-9))^(2))`
`[Cu^(2+)]=2.0xx10^(15)xx(1.0xx10^(-9))^(2)=2.0xx10^(-3)`
`:. a=2.0xx10^(-3) "mol" L^(-1)`
b. `K=([Cu^(2+)])/([Ag^(o+)]^(2)) =(2.0xx10^(-7))/((1.0xx10^(-11))^(2))=2.0xx10^(15)`
`b=2.0xx10^(15) "mol" L^(-1)`
c. `K=([Cu^(2+)])/([Ag^(o+)]^(2))`
`2.0xx10^(15)=(2.0xx10^(-2))/([Ag^(o+)]^(2))` ltbr or `[Ag^(o+)]=([2.0xx10^(-2)])/([2.0xx10^(15)])=3.16xx10^(-9)`