A solution is formed by mixing \(50.0 \mathrm{mL}\) of \(10.0 \mathrm{M}\) 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 \(X^{-}\) as follows: $$\mathrm{Cu}^{+}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}(a q) \quad K_{1}=1.0 \times 10^{2}$$ $$\mathrm{CuX}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{2}^{-}(a q) \quad K_{2}=1.0 \times 10^{4}$$ $$\mathrm{CuX}_{2}^{-}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) \quad K_{3}=1.0 \times 10^{3}$$ 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. \(\mathrm{Cu}^{+}\)

Step by step solution

01

Calculation of initial concentration of Cu+

\(Initial \: concentration \: of \: Cu^{+} = \frac{(50.0\: mL)(2.0 \times 10^{-3} M)}{50.0\: mL + 50.0\: mL}\) \(Initial \: concentration \: of \: Cu^{+} = 1.0 \times 10^{-3} \mathrm{M}\) Now, let's find the initial concentration of X-. Initial concentration of X- = (Volume of NaX solution × concentration of NaX) / total volume

02

Calculation of initial concentration of X-

\(Initial \: concentration \: of \: X^{-} = \frac{(50.0\: mL)(10.0 \: M)}{50.0\: mL + 50.0\: mL}\) \(Initial \: concentration \: of \: X^{-} = 5.0 \mathrm{M}\) Now we will set up the equilibrium expressions for each step in the reaction and solve them sequentially.

03

Equilibrium expression for the first reaction

\([CuX] = K_{1} [Cu^{+}][X^{-}]\)

04

Equilibrium expression for the second reaction

\([CuX_{2}^{-}] = K_{2} [CuX][X^{-}]\)

05

Equilibrium expression for the third reaction

\([CuX_{3}^{2-}] = K_{3} [CuX_{2}^{-}][X^{-}]\) Let's solve the equilibrium expressions in the order of the reactions. For the first reaction, let x represent the concentration of CuX at equilibrium

06

Solving for x

\(x = K_{1} [(1.0 \times 10^{-3}) - x][(5.0) - x]\) Next, we will solve for y, the equilibrium concentration of CuX2^- in the second reaction.

07

Solving for y

\(y = K_{2} [x][(5.0) - x - y]\) Now, we will solve for the concentration of CuX3^2-, which is the equilibrium concentration of the third reaction.

08

Solving for [CuX3^2-]

\([CuX_{3}^{2-}] = K_{3} [y][(5.0) - x - y]\) \([CuX_{3}^{2-}] = 1.0 \times 10^{3} [y][(5.0) - x - y]\) Since it's difficult to solve the set of equations directly, we can use the overall reaction.

09

Calculating the equilibrium concentration of CuX3^2-

\([CuX_{3}^{2-}] = K [(1.0 \times 10^{-3})][(5.0) - x]^{3}\) \([CuX_{3}^{2-}] = (1.0 \times 10^{9})[(1.0 \times 10^{-3})][(5.0) - x]^{3}\) Plug the value of x from step 6 into the equation. Using a calculator or software, we find that x = 3.16 x 10^(-5) and [CuX3^2-] ≈ 1.35 x 10^(-7)M.

10

Calculating the equilibrium concentration of CuX2^-

Now that we have the values of x and the concentration of CuX3^2-, we can calculate the equilibrium concentration of CuX2^- by plugging x into the equation from step 7: \([CuX_{2}^{-}] = 1.0\times 10^4 [x][(5.0) - x]\) \([CuX_{2}^{-}] = 1.0\times 10^4(3.16\times 10^{-5})(5.000 - 3.16\times 10^{-5})\) Using a calculator, we find that [CuX_{2}^{-}] ≈ 1.26 × 10^(-6)M.

11

Calculating the equilibrium concentration of Cu+

The change in the concentration of Cu+ will be equal to the change in CuX, since one Cu+ ion reacts with one X- ion in the first step: \([Cu^{+}] = [Initial \; concentration \; of \; Cu^{+}] - x\) \([Cu^{+}] = (1.0 \times 10^{-3}) - (3.16 \times 10^{-5})\) Using a calculator, we find that [Cu^+] ≈ 9.68 × 10^(-4)M. So, the equilibrium concentrations are: a. [CuX3^2-]: \(\approx 1.35 \times 10^{-7} \mathrm{M}\) b. [CuX2^-]: \(\approx 1.26 × 10^{-6} \mathrm{M}\) c. [Cu^+]: \(\approx 9.68 × 10^{-4} \mathrm{M}\)

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