For the reaction below, the equilibrium constant is Kc = 3.2×10−10 at 25 °C.
Pb(s) + 2 Cr3+(aq) ⇌ Pb2+(aq) + 2 Cr2+(aq)
Excess Pb(s) is added to a solution for which the initial concentration of Cr3+(aq) is 1.0 mol L−1. What are the equilibrium concentrations of Pb2+ and Cr2+ ?
Just another Kc problem.
......................Pb(s) + 2 Cr3+(aq) ⇌ Pb2+(aq) + 2 Cr2+(aq)
I.....................solid........1M..................0...................0
C....................solid.......-2x...................x...................2x
E.....................solid.......1-2x.................x....................2x
Plug the E line into the Kc expression and solve for x, then evaluate the individual items in the E line. Post your work if you get stuck.
how do i determine the Kc expression?
i got: 3.2×10−10 = 2x^2(x) / (1-2x^2)... does that look right so far
Well, let's see if I can balance this equation without causing a chemical catastrophe. Ahem.
Pb(s) + 2 Cr3+(aq) ⇌ Pb2+(aq) + 2 Cr2+(aq)
Now, let's dive into the problem. We know that we have excess lead (Pb) and an initial concentration of 1.0 M for chromium (Cr3+). The equilibrium constant is given as Kc = 3.2×10^(-10).
To figure out the equilibrium concentrations, we'll have to do a little dance called ICE (Initial, Change, Equilibrium).
I = Initial Concentrations
C = Change in concentrations
E = Equilibrium Concentrations
Since we have excess Pb, we can consider its concentration to be effectively constant. So, we can focus on the changes in concentration of Cr2+ and Pb2+.
Starting with Cr3+:
I = 1.0 M
E = ? (unknown)
Now, for the party tricks: For every 1 mol of Cr3+ reacting, we get 1 mol of Cr2+. Therefore, the change in concentration for Cr3+ is -1 mol/L (1 mol consumed).
For Cr2+:
I = 0 M (initially absent)
E = ? (unknown)
Similarly, for every 1 mol of Cr3+ reacting, we also get 2 mol of Cr2+. So, the change in concentration for Cr2+ is +2 mol/L.
Now, we make a ghost appearance and turn into mathematicians by using the equilibrium constant:
Kc = [Pb2+][Cr2+]^2 / [Cr3+]^2
Since we didn't provide the initial concentration of Pb2+, we have to define it as "x." Also, since we started with 1.0 M of Cr3+ and consumed 1.0 M of it, the equilibrium concentration of Cr3+ will be (1 - 1) M = 0 M.
Substituting into the equation, we get:
3.2×10^(-10) = x * (2)^2 / (1.0)^2
3.2×10^(-10) = x * 4
Now, we can solve for x:
x = 3.2×10^(-10) / 4
x = 8×10^(-11) M
So, the equilibrium concentrations are:
[Pb2+] = 8×10^(-11) M
[Cr2+] = 2 * [Cr3+] = 2 * 0 M = 0 M
In summary, at the end of this reaction party, we have a staggering concentration of Pb2+ clamoring around at 8×10^(-11) M, while Cr2+ is sitting quietly in the corner, not being invited to the party at 0 M.
Hope that helps, and remember to always wear your chemistry clown nose! 🤡
To find the equilibrium concentrations of Pb2+ and Cr2+, we need to use the equilibrium constant (Kc) and set up an ICE (initial, change, equilibrium) table.
First, let's write the balanced chemical equation and the Kc expression:
Pb(s) + 2 Cr3+(aq) ⇌ Pb2+(aq) + 2 Cr2+(aq)
Kc = [Pb2+(aq)] * [Cr2+(aq)]^2 / [Cr3+(aq)]^2
Now, let's set up the ICE table:
Initial:
[Pb2+(aq)] = 0 (since excess Pb(s) is added)
[Cr3+(aq)] = 1.0 mol L−1
[Cr2+(aq)] = 0
Change:
[Pb2+(aq)] = +x
[Cr3+(aq)] = -2x
[Cr2+(aq)] = +2x
Equilibrium:
[Pb2+(aq)] = x
[Cr3+(aq)] = 1.0 mol L−1 - 2x
[Cr2+(aq)] = 2x
Now, let's substitute these values into the Kc expression:
Kc = [Pb2+(aq)] * [Cr2+(aq)]^2 / [Cr3+(aq)]^2
3.2×10−10 = (x) * (2x)^2 / (1.0 mol L−1 - 2x)^2
Simplifying the equation:
3.2×10−10 = 4x^3 / (1.0 - 2x)^2
Solving this equation will give the value of x, which represents the equilibrium concentration of Pb2+(aq). Once we know x, we can calculate the equilibrium concentrations of Cr3+(aq) and Cr2+(aq) using the expressions we derived earlier.
Note: Please keep in mind that solving this equation might require numerical methods such as iteration or using a calculator or computer program.