Question & Answer: It is given that, [NH3]0 [H+]0 = [NH3]x [H+]x (equation 1) where subscript 0 refers to the…..

It is given that, [NH₃]₀ [H⁺]₀ = [NH₃]_x [H⁺]_x (equation 1) where subscript 0 refers to the initial values in a stock solution of NH₃, and the subscript x refers to the values in an unknown mixture (i.e during titration) both in 2M NH₄NO₃. Since pX is defined as the negative log₁₀ of X, equation (1) can be written as:

p(NH₃)₀ + pH₀ = p(NH₃)_x + pH_x (equation 2)

p(NH₃)_x = p(NH₃)₀ + pH₀ – pH_x (equation 3) where pH_x is the measured pH at any time during titration. Thus, concentration of free NH₃ can be calculated directly by applying equation 3. The total Cu²⁺ in solution at any point is defined as c_cu where

c_cu = [Cu²⁺] + [Cu(NH₃)²⁺] [Cu(NH₃)₂²⁺] + [Cu(NH₃)₃²⁺] + [Cu(NH₃)₄²⁺​]

The fractions, a_n of each species present at any point can be found by using the expressions for the stepwise formation constants.

a₀ = [M]/c_cu = (1+ K₁[NH₃] + K₁K₂[NH₃]² + K₁K₂K₃[NH₃]³)^-1

a₁ = [Cu(NH₃)²⁺] / c_cu = K₁[NH₃]a₀

a₂ = [Cu(NH₃)₂²⁺] / c_cu = K₁K₂[NH₃]²a₀

The average number of NH₃ bound per Cu²⁺ or the ligand number (n_av) can be calculated:

n_av = [NH₃]_bound / c_cu = ([NH₃]_total – [NH₃]_free) / c_cu = a₁+2a₂₊3a₃+4a₄

In the experiment, the 0.7495 g (3 mmol) copper sulfate was dissolved in 100mL 2M NH₄NO₃ solution to form 0.030 M Cu²⁺ solution (100 mL). 50 mL of copper solution was transferred to 200 mL beaker and titrated with 2M NH₄NO₃ solution from a 50mL burette in 2 mL aliquots. The pH was measured after each addition. Given the information, calculate

Calculate Moles of ammonia added (mmol), p(NH₃), [NH₃] calculated from p(NH₃), Moles of free ammonia (mmol), Moles bound ammonia (mmol), N_av

If the Moles of ammonia added (mmol) cannot be calculated, pretend that 1st value of moles of ammonia added was 0.384 mmol.

Thank you very much