Lead metal is added to 0.140 Cr^3+(aq).
Pb(s)+ 2Cr^3(aq)<---> Pb^2+(aq)+2Cr^2+(aq)
KC=3.2*10^{-10}
A. What is Pb^2+ when equilibrium is established in the reaction?
B. What is Cr^2+ when equilibrium is established in the reaction?
C. What is Cr^3+ when equilibrium is established in the reaction?
2 answers
* correction 0.140 M Cr^3+(aq).
You didn't make a correction with your correction.
Pb(s) + 2Cr^+3 ==> 2Cr^+2 + Pb^+2
Do an ICE chart.
Initial:
Cr^+3 = 0.140
Cr+2 = 0
Pb^+2 = 0
Change:
Cr^+3 = -2y
Cr^+2 = +2y
Pb^+2 = +y
Equilibrium:
Cr^+2 = +2y
Pb^+2 = +y
Cr^+3 = 0.140-2y
Plug all of that into the expression for Kc and solve for y. I think it's a cubic equation.
Pb(s) + 2Cr^+3 ==> 2Cr^+2 + Pb^+2
Do an ICE chart.
Initial:
Cr^+3 = 0.140
Cr+2 = 0
Pb^+2 = 0
Change:
Cr^+3 = -2y
Cr^+2 = +2y
Pb^+2 = +y
Equilibrium:
Cr^+2 = +2y
Pb^+2 = +y
Cr^+3 = 0.140-2y
Plug all of that into the expression for Kc and solve for y. I think it's a cubic equation.
1 answer