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?
* 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.
To answer these questions, we need to use the concept of equilibrium and the equilibrium constant (Kc) of the reaction. The equilibrium constant expresses the ratio of products to reactants at equilibrium, and it is determined by the concentrations of the species involved in the reaction.
In this case, we have the equation:
Pb(s) + 2Cr^3+(aq) ⇌ Pb^2+(aq) + 2Cr^2+(aq)
The equilibrium constant (Kc) for this reaction is given as 3.2 x 10^(-10), which is the ratio of the concentration of products to the concentration of reactants.
Now let's answer each question:
A. What is Pb^2+ when equilibrium is established in the reaction?
To find the concentration of Pb^2+ at equilibrium, let's assume x represents the concentration of Pb^2+. Since the stoichiometry of the reaction is 1:1 between Pb^2+ and Pb(s), the concentration of Pb^2+ will be equal to the concentration of Pb(s) at equilibrium.
Therefore, [Pb^2+] = [Pb(s)] = x
B. What is Cr^2+ when equilibrium is established in the reaction?
Similar to the previous question, we can assume y represents the concentration of Cr^2+. Since the stoichiometry of the reaction is 2:1 between Cr^2+ and Cr^3+, the concentration of Cr^2+ will be twice the concentration of Cr^3+ at equilibrium.
Therefore, [Cr^2+] = 2[Cr^3+] = 2y
C. What is Cr^3+ when equilibrium is established in the reaction?
To determine the concentration of Cr^3+, we need to use the equilibrium constant (Kc). The expression for Kc is given by:
Kc = ([Pb^2+][Cr^2+]^2) / ([Cr^3+]^2)
Substituting the values we obtained for [Pb^2+] and [Cr^2+] in terms of x and y:
Kc = (x)(2y)^2 / (y^2)
Now substitute the given value of Kc into the equation and solve for [Cr^3+]:
3.2 x 10^(-10) = (x)(4y^2) / (y^2)
Simplifying the equation, we get:
3.2 x 10^(-10) = 4x
Solve for x to find the concentration of Cr^3+ at equilibrium.
These calculations will give you the concentration of Pb^2+, Cr^2+, and Cr^3+ when equilibrium is established in the reaction.