Question

A) A student mixes in a test tube 3.00mL of 0.050M CuSO4 with 7.00mL of 0.20M...

A) A student mixes in a test tube 3.00mL of 0.050M CuSO4 with 7.00mL of 0.20M NH3/NH41+. The solution becomes a deep blue color. Assuming all the Cu2+ is complexed with NH3 to form the [Cu(NH3)4]2+ ion, determine the concentration of the complex in the solution.

B) The standard solution made in question #1 was then scanned and the wavelength of maximum absorbance was determined. At the wavelength of maximum absorbance (minimum %T) the %T was found to be 12.16%. Using the concentration from question #1and the calculated absorbance, determine εl for the standard solution.

C) Test tube #1 was prepared as described in the procedure. Test tube #1 solution was placed in the spectrophotometer and at the wavelength of maximum absorbance the %T wasfound to be 97.05%. Using εl form question 2 and the calculated absorbance of test tube#1, determine the [Cu(NH3)4]2+ ion in the solution.

D) Calculate the [NH3]I initially and [NH3]E at equilibrium. (The concentration of [NH3]E will be the initial [NH3] minus the amount complexed to the copper according to the balanced overall equation.)

E) Using the values for Ksp and Kb, calculate the [Cu2+].

F) Calculate Kf for the complex using the [Cu(NH3)4]2+ found in question 3, [NH3]E found in question 4, and [Cu2+] found in question 5.

Homework Answers

Answer #1

Part A.

The no. of millimoles of Cu2+ = 3 mL * 0.05 mmol/mL = 0.15 mmol

The total volume = (3+7) mL = 10 mL

Therefore, the conentration of the complex in the solution (c) = 0.15 mmol/10 mL = 0.015 M

Part B.

Absorbance (A) = -Log(T)

i.e. A = -Log(12.16/100)

i.e. A = 0.915

According to Beer-Lambert's law: A = cl

i.e. 0.915 = * 0.015 M * l

i.e. l = 0.915/0.015 M = 61 M-1

Part C.

A = -Log(97.05/100) = 0.013

Now, 0.013 = 61 M-1 * c

Therefore, [Cu(NH3)4]2+ (c) = 0.013/61 M-1 = 2.132*10-4 M

Part D.

[NH3]I = 7*0.2 mmol/10 mL = 0.14 M

[NH3] consumed = 0.015*4 = 0.06 M

Therefore, [NH3]E = 0.14 - 0.06 = 0.08 M

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