Question

please help!!! please show work as i am so confused!!

U U DUVJUJU. I LILI PIDU 01 12. 14. The following reaction produces the deep blue tetraamminecopper(II) complex ion: Cu? (aq) + 4NH3(aq) 5 Cu(NH3)4** (aq) To determine the equilibrium constant for the reaction above, an equilibrium solution was prepared by mixing together 10.00 mL of a 0.0750 M Cu(NO)solution with 5.00 mL of a 0.600 M NH, solution The absorbance of the equilibrium solution was measured at a wavelength of 546 nm in a 1.00 cm cuvet, and found to be 0.356. (a) Determine the initial concentrations of Beer's Law Graph Cu?* and NH3 in the equilibrium solution. (b) Using the calibration line to the right, Abs = mx + b determine the equilibrium concentration m(slope): 9.25 of Cu(NH3)** in the equilibrium bly-intercept): 0.002 solution (c) Determine the equilibrium concentrations of Cu?* and NH, in the equilibrium solution 0.0 (d) Determine the equilibrium constant for 0.00 0.02 0.04 0.06 0.08 0.10 0.12 the given reaction. M Cu(NH) . Absorbance

Answer 1

(a) Use the dilution equation.

M₁V₁ = M₁V₂

where the symbols have their usual meanings.

Total volume, V₂ = (10.00 + 5.00) mL = 15.00 mL.

Use the above equation for Cu²⁺ and NH₃.

(10.00 mL)*(0.0750 M) = (15.00 mL)*[Cu²⁺]_i

=======> [Cu²⁺]_i = (10.00 mL)*(0.0750 M)/(15.00 mL) = 0.050 M (ans).

(5.00 mL)*(0.600 M) = (15.00 mL)*[NH₃]_i

=======> [NH₃]_i = (5.00 mL)*(0.600 M)/(15.00 mL) = 0.200 M (ans).

(b) Use the regression equation with m = 9.25, b = 0.002 and A = 0.356.

Plug in values in the regression equation and get

0.356 = (9.25)*[Cu(NH₃)₄²⁺] + 0.002

where [Cu(NH₃)₄²⁺] denotes the molar concentration of Cu(NH₃)₄²⁺ at equilibrium.

[Cu(NH₃)₄²⁺] = (0.356 – 0.002)/9.25 = 0.0383 ≈ 0.038 (ans, correct to 2 sig. figs).

The equilibrium concentration of Cu(NH₃)₄²⁺ is 0.038 M.

(c) [Cu²⁺]_eq = [Cu²⁺]_i – [Cu(NH₃)₄²⁺]_eq

= (0.050 M) – (0.038 M)

= 0.012 M.

[NH₃]_eq = [NH₃]_i – 4*[Cu(NH₃)₄²⁺]_eq (1 mol Cu(NH₃)₄²⁺ requires 4 moles NH₃).

= (0.200 M) – 4*(0.038 M)

= 0.048 M.

Write down the expression for the equilibrium constant as

K = [Cu(NH₃)₄²⁺]/[Cu²⁺][NH₃]⁴

where [..] denote equilibrium concentrations

= (0.038 M)/(0.012)(0.048 M)⁴

= 5.96*10⁵ (ignore units)

≈ 6.0*10⁵ (ans).