Question

A solution is formed by mixing \(50.0 \mathrm{~mL}\) of \(10.0 \mathrm{M} \mathrm{NaX}\) with \(50.0 \mathrm{~mL}\) of \(2.0 \times 10^{-3} \mathrm{M} \mathrm{CuNO}_{3}\). Assume that \(\mathrm{Cu}^{+}\) forms complex ions with \(\mathrm{X}^{-}\) as follows: $$ \begin{aligned} \mathrm{Cu}^{+}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \operatorname{CuX}(a q) & K_{1}=1.0 \times 10^{2} \\ \mathrm{CuX}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{2}-(a q) & K_{2}=1.0 \times 10^{4} \\ \mathrm{CuX}_{2}^{-}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}{ }^{2-}(a q) & K_{3}=1.0 \times 10^{3} \end{aligned} $$ with an overall reaction \(\mathrm{Cu}^{+}(a q)+3 \mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) \quad K=1.0 \times 10^{9}\) Calculate the following concentrations at equilibrium. a. \(\mathrm{CuX}_{3}^{2-}\) b. \(\mathrm{CuX}_{2}^{-}\) c. \(\mathrm{Cu}^{+}\)

Short answer

The equilibrium concentrations are: a. \([CuX_{3}^{2-}] \approx 1.25 \times 10^{-7} \mathrm{M}\) b. \([CuX_{2}^{-}] \approx 1.25 \times 10^{-7} \mathrm{M}\) c. \([Cu^+] \approx 1.0 \times 10^{-3} \mathrm{M}\)

Step-by-step solution

List the initial concentrations

Before we start calculating the equilibrium concentrations, let's note down the initial concentrations of Cu+ and X- ions. We know that we have 50.0 mL of 10.0 M NaX and 50.0 mL of 2.0x10^{-3} M CuNO3. Since the volume of the solution doubled after mixing, the initial concentrations are: - [Cu+] = (50.0 mL x 2.0x10^{-3} M) / 100.0 mL = 1.0x10^{-3} M - [X-] = (50.0 mL x 10.0 M) / 100.0 mL = 5.0 M

Calculate the equilibrium concentration of CuX32-

Let's use 'x' for the equilibrium concentration of CuX32-. Since the overall reaction is: Cu+ + 3X- ⇌ CuX32-, we can write the equilibrium expression: \[K = \frac{[CuX_{3}^{2-}]}{[Cu^+][X^-]^3}\] We know K=1.0x10^9, [Cu+] = 1.0x10^{-3}-x, and [X-] = 5-x. Plugging these values into the expression: \[1.0 \times 10^9 = \frac{x}{(1.0 \times 10^{-3}-x)(5-x)^3}\] First, we will estimate if x is extremely small compared to 1.0x10^{-3} and 5 for simplification purposes. If that's the case, x would not considerably affect the concentrations. So, let's assume x is negligible and ignore it in the equation: \[1.0 \times 10^9 = \frac{x}{(1.0 \times 10^{-3})(5)^3}\] Now we can solve for x: \[x \approx 1.25 \times 10^{-7} \mathrm{M}\] And since x is indeed negligibly small, we conclude that the equilibrium concentration of CuX32- is approximately: \[ [CuX_{3}^{2-}] \approx 1.25 \times 10^{-7} \mathrm{M}\]

Calculate the equilibrium concentration of CuX2-

We can use the K2 value to calculate the concentration of CuX2-: \[K_2 = \frac{[CuX_{2}^{-}]}{[CuX][X^-]}\] The concentration of CuX is difficult to find, but we can use another relationship instead: Since CuX32- is formed from CuX2-, there is 1:1 stoichiometry (From Reaction 2 in the problem statement: CuX(aq) + X-(aq) ⇌ CuX2-(aq)). Thus, we can assume [CuX2-] = [CuX32-] = x: \[ [CuX_{2}^{-}] = 1.25 \times 10^{-7} \mathrm{M}\]

Calculate the equilibrium concentration of Cu+

To find the equilibrium concentration of Cu+, we can make use of the fact that the initial concentration of Cu+ has been reduced by the formation of CuX32-. We previously assumed x ([CuX32-]) is negligibly small compared to the initial Cu+ concentration, but we can still account for it: \[ [Cu^+] = [Cu^+]_{initial} - 3[ CuX_{3}^{2-} ]\] \[ [Cu^+] = 1.0 \times 10^{-3} \mathrm{M} - 3(1.25 \times 10^{-7} \mathrm{M}) \] \[ [Cu^+] \approx 1.0 \times 10^{-3} \mathrm{M}\] So, the equilibrium concentrations are: a. [CuX32-] ≈ 1.25x10^{-7} M b. [CuX2-] ≈ 1.25x10^{-7} M c. [Cu+] ≈ 1.0x10^{-3} M

Key concepts

Chemical Equilibrium

Imagine a seesaw perfectly balanced in the middle, not tipping to either side, regardless of the ups and downs. This is similar to what happens in a chemical reaction when it reaches a state of chemical equilibrium. In this dynamic state, the rate of the forward reaction (where reactants are converted to products) is equal to the rate of the reverse reaction (where products revert to reactants).

However, reaching equilibrium does not mean that the reactants and products are present in equal amounts. It simply indicates that their concentrations no longer change with time. The seesaw analogy helps visualize the balance of the process, not the amounts of reactants or products.

Equilibrium Constants

To quantify the point at which a chemical system is at equilibrium, scientists use a mathematical expression known as the equilibrium constant, represented by the letter K. This expression involves the concentrations of the products and reactants raised to the power of their coefficients from the balanced chemical equation. The value of K offers insight into the proportions of reactants and products present when a system is at equilibrium.

The larger the value of K, the more product-favored the reaction is. On the other hand, a small K value indicates that the reactants are more predominant at equilibrium. The equilibrium constant is crucial in predicting the position of equilibrium and understanding how changes in conditions, like concentration or temperature, can affect the reaction.

Reaction Stoichiometry

The term stoichiometry stems from the Greek words 'stoicheion' (element) and 'metron' (measure), and in the realm of chemistry, it draws up a map that navigates the quantities of reactants and products in a chemical reaction. It is founded on the law of conservation of mass where the number of atoms of each element must be preserved throughout the reaction.

Importance of Stoichiometry in Equilibrium

Stoichiometry is vital while calculating equilibrium concentrations because it gives us the ratios in which reactants combine and products form. In our problem, reaction stoichiometry allows us to use the concentration of one species to find another. Understanding these ratios is like having a recipe at hand; if you know the amount of one ingredient, you can determine the amounts of the others needed to complete the dish.

Complex Ion Formation

When a metal ion binds with one or more ligands - molecules or ions that can donate a pair of electrons - a complex ion is formed. This is like a dance, where the metal ion is the lead partner, and the ligands follow along. In the given problem, \r\(\text{Cu}^+\) pairs up with \r\(\text{X}^-\) ligands to form a series of complex ions: \r\(\text{CuX}, \text{CuX}_2^-\), and \r\(\text{CuX}_3^{2-}\).\r

Each forward step toward more complex structures - adding more ligands - is characterized by a corresponding equilibrium constant (\r\(K_1, K_2, K_3\)). The formation of these complexes not only depends on the stoichiometry but also the inherent tendency of the metal ion to bind with ligands, quantified by these equilibrium constants. In a concentrated soup of metal ions and ligands, knowing how to calculate the formation of complex ions helps chemists predict which complexes will predominate and in what concentration.