Lead metal is added to 0.150 M Cr3+(aq).
Pb(s)+2Cr3+(aq)⇌Pb2+(aq)+2Cr2+(aq).
KC=3.2×10−10
What is [Pb2+] when equilibrium is established in the reaction?
What is [Cr2+] when equilibrium is established in the reaction?
What is [Cr3+] when equilibrium is established in the reaction?
See my response above.
To determine the equilibrium concentrations of [Pb2+], [Cr2+], and [Cr3+], we can use the equilibrium constant expression (Kc) and the stoichiometric coefficients from the balanced equation.
The equilibrium constant expression for the given reaction is:
Kc = ([Pb2+][Cr2+]^2) / ([Cr3+]^2)
Given that Kc = 3.2×10^−10 and the initial concentration of Cr3+ is 0.150 M, we can now solve for the equilibrium concentrations of [Pb2+], [Cr2+], and [Cr3+].
To find [Pb2+]:
Since the stoichiometric coefficient for Pb2+ is 1, when equilibrium is established, the concentration of Pb2+ will also be equal to the concentration of lead metal that was added initially. Therefore, [Pb2+] = 0.150 M.
To find [Cr2+]:
The stoichiometric coefficient for Cr2+ is 2, meaning that for every 2 moles of Cr2+ formed, we will have consumed 2 moles of Cr3+. Thus, the concentration of Cr2+ will be twice the change in [Cr3+].
Given that the initial concentration of Cr3+ is 0.150 M, let's assume x is the change in [Cr3+] at equilibrium. Therefore, the concentration of Cr2+ will be 2x. We need to solve for x to find the value of [Cr2+].
Substituting into the equilibrium constant expression, we have:
3.2×10^−10 = (0.150)(2x)^2 / (x)^2
Simplifying the equation:
3.2×10^−10 = (0.150)(4x^2) / x^2
3.2×10^−10 = 0.600
4x^2 = 0.600
Solving for x:
x^2 = 0.150
x ≈ 0.387 M
Since the change in [Cr3+] is approximately 0.387 M, the concentration of [Cr2+] at equilibrium is 2(0.387) = 0.774 M.
To find [Cr3+]:
The initial concentration of Cr3+ was 0.150 M, and the change in concentration at equilibrium is approximately 0.387 M. Thus, the concentration of Cr3+ at equilibrium can be calculated as:
[Cr3+] = initial concentration - change in concentration
[Cr3+] = 0.150 - 0.387
[Cr3+] ≈ -0.237 M
Since concentrations can't be negative, we can conclude that [Cr3+] is approximately 0 M at equilibrium.
Therefore, the equilibrium concentrations are:
[Pb2+] = 0.150 M
[Cr2+] = 0.774 M
[Cr3+] ≈ 0 M