Initial concentration of Cu2+ = ( 4.6 ×10-3mol/0.470L) × 1L = 9.79×10-3M
Initial concentration of NH3 = 0.39M
consider the complete formation of Cu(NH3)42+
Cu2+(aq) + 4NH3(aq) ----------> Cu(NH3)42+
0.00979 M of Cu2+ react with 0.03916M of NH3 to give 0.00979M of Cu(NH3)42+
remaining concentration of NH3 = 0.39M - 0.03916M = 0.3508M
Now , consider the dissociation equillibrium of Cu(NH3)42+
Cu(NH3)42+ (aq) <-------> Cu2+(aq) + 4NH3(aq)
Kd = [Cu2+] [NH3]4/[Cu(NH3)42+]
Kd = 1/Kf = 1/5.6 ×1011= 1.79×10-12
Initial concentration
[Cu(NH3)42+] = 0.00979
[Cu2+] = 0
[NH3] = 0.3508
change in comcentration
[ Cu(NH3)42+] = -x
[Cu2+]= +x
[NH3] = + 4x
equillibrium concentration
[Cu(NH3)42+] = 0.00979 - x
[Cu2+] = x
[NH3] = 0.3508 + 4x
so,
x(0.3508 + 4x )4 / (0.0079 - x) = 1.79 ×10-12
we can assume 0.3508 + 4x = 0.3508 and 0.0079 - x = 0.0079
x(0.3508)4/(0.0079) = 1.79 ×10-12
x 1.917 = 1.79×10-12
x = 9.34 × 10-13
Therefore, at equillibrium
concentration of Cu2+ = 9.34×10-13M
The complex ion Cu(NH3)42+ is formed in a solution made of 0.0100 M Cu(NO3)2 and 0.300 M NH3. What are the concentrations of Cu2 , NH3, and Cu(NH3)42+ at equilibrium? The formation constant*, Kf, of Cu(NH3)42+ is 1.70 × 1013
The complex ion Cu(NH3)42+ is formed in a solution made of 0.0300 M Cu(NO3)2 and 0.400 M NH3. What are the concentrations of Cu2+, NH3, and Cu(NH3)42+ at equilibrium? The formation constant*, Kf, of Cu(NH3)42+ is 1.70 x 1013 Suppose you have a solution that contains 0.0440 M Ca2+ and 0.0980 M Ag+. If solid Na3PO4 is added to this mixture, which of the following phosphate species would precipitate out of solution first? When the second cation just starts to precipitate, what percentage...
In the presence of excess OH-, the Zn2+(aq) ion forms a hydroxide complex ion, Zn(OH)42-. Calculate the concentration of free Zn2+ ion when 1.54×10-2 mol Zn(CH3COO)2(s) is added to 1.00 L of solution in which [OH- ] is held constant (buffered at pH 12.70). For Zn(OH)42-, Kf = 4.6×1017. [Zn2+] = M
1- In the presence of excess OH-, the Zn2+(aq) ion forms a hydroxide complex ion, Zn(OH)42-. Calculate the concentration of free Zn2+ ion when 1.32×10-2mol ZnSO4(s) is added to 1.00 L of solution in which [OH- ] is held constant (buffered at pH 12.40). For Zn(OH)42-, Kf = 4.6×1017. [Zn2+] = ------ M 2- What is the approximate concentration of free Hg2+ ion at equilibrium when 1.86×10-2 mol mercury(II) nitrate is added to 1.00 L of solution that is 1.310...
In the presence of excess OH-, the Zn2+(aq) ion forms a hydroxide complex ion, Zn(OH)42-. Calculate the concentration of free Zn2+ ion when 1.57×10-2 mol Zn(NO3)2(s) is added to 1.00 L of solution in which [OH- ] is held constant (buffered at pH 12.20). For Zn(OH)42-, Kf = 4.6×1017. [Zn2+]=___M please explain
Silver (I) ions in aqueous solutions react with NH3(aq) to form a complex ion according to the following reaction: Ag+ + 2NH3(aq) ---> Ag(NH3)2+ with a formation constant, Kf of 1.5 x 107. Calculate the solubility in g/L, of AgCN(s) (Ksp = 2.2 x 10-16) in 0.75M NH3(aq)
A metal ion M forms a stable complex with the ligand X: M + (aq) + 4 X (aq) ⇌ M (X) 4+ (aq) Kf = 1.0 ∙ 10^20 In a solution where [M +] = 0.100 M and [X] = 0.800 M before reaction, what is [X] at equilibrium?
please help What is the approximate concentration of free Cu* ion at equilibrium when 1.57x102 mol copperII) nitrate is added to 1.00 L of solution that is 1.380 M in NH3. For [Cu(NH3)4j2, Kf 2.1 1013 [Cu2 м What is the approximate concentration of free Fe2* ion at equilibrium when 1.42x102mol iron(II) nitrate is added to 1.00 L of solution that is 1.38so M in CN. For [Fe(CN)6]4, Kf=1.0x1035 [Fe"]= In the presence of excess OH, the Al*(aq) ion forms...
In the presence of excess OH-, the Al3+(aq) ion forms a hydroxide complex ion, Al(OH)4-. Calculate the concentration of free Al3+ ion when 1.32×10-2 mol Al(NO3)3(s) is added to 1.00 L of solution in which [OH- ] is held constant (buffered at pH 13.00). For Al(OH)4-, Kf = 1.1×1033. [Al3+] = ?M
Cu(OH)2 has Ksp = 2.2 x 10^–20. The complex ion Cu(NH3)4 2+ has Kf = 2.1 x 10^13 . What concentration of [NH3] can dissolve 0.01mol/L Cu(OH)2 to become Cu(NH3)4 2+ in 1L solution?