Question

1. lead metal is added to 0.100 m cr³⁺(aq). what are pb²⁺, cr²⁺, and cr³⁺ when the following equilibrium is established? kₑ = 3.2 × 10¹⁰

pb(s) + 2 cr³⁺(aq) ⇌ pb²⁺(aq) + 2 cr²⁺(aq)

Explanation:

Step1: Set up the ICE table

Let the change in concentration of $Cr^{3 + }$ be $2x$.

	$Cr^{3+}$	$Pb^{2+}$	$Cr^{2+}$
Change (M)	- 2x	+ x	+ 2x
Equilibrium (M)	0.100 - 2x	x	2x

Step2: Write the equilibrium - constant expression

The equilibrium - constant expression for the reaction $Pb(s)+2Cr^{3 + }(aq)
ightleftharpoons Pb^{2 + }(aq)+2Cr^{2 + }(aq)$ is $K_{c}=\frac{[Pb^{2 + }][Cr^{2 + }]^{2}}{[Cr^{3 + }]^{2}}$.
Substitute the equilibrium - concentrations into the expression: $K_{c}=\frac{x(2x)^{2}}{(0.100 - 2x)^{2}}$. Since $K_{c}=3.2\times10^{10}$ is very large, the reaction proceeds far to the right. We can assume that $0.100-2x\approx0.100$ (because x is very small compared to 0.100).

Step3: Solve for x

The equilibrium - constant expression becomes $3.2\times10^{10}=\frac{x(2x)^{2}}{(0.100)^{2}}$.
First, simplify the right - hand side: $\frac{x(2x)^{2}}{(0.100)^{2}}=\frac{4x^{3}}{0.0100}$.
Then, solve for x:
\[

$$\begin{align*} 3.2\times10^{10}&=\frac{4x^{3}}{0.0100}\\ 4x^{3}&=3.2\times10^{10}\times0.0100\\ 4x^{3}&=3.2\times10^{8}\\ x^{3}& = 8\times10^{7}\\ x&=431. \end{align*}$$

\]
But our assumption that $0.100 - 2x\approx0.100$ was wrong. Let's solve the equation $3.2\times10^{10}=\frac{x(2x)^{2}}{(0.100 - 2x)^{2}}$ without approximation.
\[

$$\begin{align*} 3.2\times10^{10}&=\frac{4x^{3}}{(0.100 - 2x)^{2}}\\ 3.2\times10^{10}(0.100 - 2x)^{2}&=4x^{3}\\ 3.2\times10^{10}(0.0100 - 0.4x+4x^{2})&=4x^{3}\\ 3.2\times10^{8}-1.28\times10^{10}x + 1.28\times10^{11}x^{2}&=4x^{3}\\ 4x^{3}-1.28\times10^{11}x^{2}+1.28\times10^{10}x - 3.2\times10^{8}&=0 \end{align*}$$

\]
Since $K_{c}$ is large, the reaction goes almost to completion. Let's assume that all of the $Cr^{3 + }$ is consumed.
If all of the $Cr^{3 + }$ is consumed, from the stoichiometry of the reaction, when $[Cr^{3 + }]=0$, for every 2 moles of $Cr^{3 + }$ consumed, 1 mole of $Pb^{2 + }$ and 2 moles of $Cr^{2 + }$ are produced.
Initial $n_{Cr^{3+}}=0.100$ M. When the reaction goes to completion, $[Pb^{2 + }]=\frac{0.100}{2}=0.050$ M, $[Cr^{2 + }]=0.100$ M, and $[Cr^{3 + }]\approx0$ M.

Answer:

$[Pb^{2 + }]=0.050$ M, $[Cr^{2 + }]=0.100$ M, $[Cr^{3 + }]\approx0$ M