Question: Solid silver is added to a solution with these initialconcentrations:

Question

Question:

Solid silver is added to a solution with these initialconcentrations:

[Ag]+ = 0.200M

[Fe]2+ = 0.100M

[Fe]3+ = 0.300M

The following revesible reaction occurs:

Ag+(aq)   +  Fe2+(aq)  <=====> Ag(s) + Fe3+(aq)           Keq = 2.98

What are the ion concentrations when equilibrium isestablished?

(Hint – use Q, ICE, Quad)

My start to the solution:

Balanced equation:  Ag+(aq)   +  Fe2+(aq)  <=====> Ag(s) + Fe3+(aq)           

Equilibrium expression:    k = 2.98 = ([Ag][Fe3+] ) / ( [Ag+][Fe2+] )

Initial Concentrations:   [Ag]+ = 0.200M

[Fe]2+ = 0.100M

[Fe]3+ = 0.300M

[Not sure how to calculate Q and following ofthis problem, any help would be great!!!]

Explanation / Answer

make an ICE table       Ag+(aq)  +   Fe2+(aq)  <=====> Ag(s) + Fe3+(aq) I      0.2               0.1                                       0.3 C      +x               +x                                          -x E 0.2+x            0.1+x                                    0.3-x and kc = [Fe3+]/[Ag+][Fe2+] = [0.3-x]/[0.2+x][0.1+x] you dont put [Ags] in your equation because its insolid state and it doesn't effect the concentrations of the ions.so you just ignore the solid silver. use the quadratic formula to solve the above equation and youget x to be equal to x=0.11. so, to obtain equilibrium concentrations Ag = 0.2 + 0.11= 0.31 Fe2+ = 0.1 + 0.11= 0.21 Fe3+ = 0.3 - 0.11 = 0.19 Q is used to tell which direction the reaction Q = [Fe3+]/[Ag+][Fe2+] Q = 0.19 / 0.31 x 0.21 = 2.9 since Q is almost equal to Kc, it means the reaction is atequilbrium. HOpe that helps SORRY ABOUT THE DOUBLE POSTS, IT WAS AN ACCIDENT.

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