Zinc (II) ions form a strong complex in the presence of EDTA. Calculate the equilibrium concentration...
Zinc (II) ions form a strong complex in the presence of EDTA. Calculate the equilibrium concentration of Zn2+ and Zn(EDTA)2- after 0.600 mol of Zn(NO3)2has been dissolved in 1.00L of 2.60 M EDTA solution. The Kf of Zn(EDTA)2- is 3.8*1016.
Is there a way you can do it without using the quadratic equation?
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Zinc (II) ions form a strong complex in the presence of EDTA.
Calculate the equilibrium concentration...
Zinc(II) ions forms a strong complex in the presence of EDTA. Calculate the equilibrium concentrations of...
Zinc(II) ions forms a strong complex in the presence of EDTA. Calculate the equilibrium concentrations of Zn2+ and Zn(EDTA)2- after 0.600 mol of Zn(NO3)2 has been dissolved in 1.00 L of 2.60 M EDTA solution. The Kf of Zn(EDTA)2- is 3.8 x 1016.
Quest. 6 (20 pts). Nickel (II) ions forms a strong complex in the presence of ammonia...
Quest. 6 (20 pts). Nickel (II) ions forms a strong complex in the presence of ammonia (NH). Calculate the equilibrium concentrations of NP and [Ni(NHLP complex after 0.120 mol of Ni(NO3)2 has been dissolved in 1.00 L of 2.30 M NH, solution. The Kiof (Ni(NHSP is 2.0 x 10. N" (aq) +6 NHỊ (aq) # [NINH " (aq)
A) Consider the insoluble compound zinc hydroxide, Zn(OH)2. The zinc ion also forms a complex with...
A) Consider the insoluble compound zinc hydroxide, Zn(OH)2. The zinc ion also forms a complex with ammonia. Write a balanced net ionic equation to show why the solubility of Zn(OH)2(s) increases in the presence of ammonia and calculate the equilibrium constant for this reaction. For Zn(NH3)42+, Kf = 2.9×109. Specify states such as (aq) or (s) and provide K. K = ______ B) Consider the insoluble compound nickel(II) hydroxide, Ni(OH)2. The nickel ion also forms a complex with cyanide ions....
In the presence of excess OH-, the Zn2+(aq) ion forms a hydroxide complex ion, Zn(OH)42-. Calculate...
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
1- In the presence of excess OH-, the Zn2+(aq) ion forms a hydroxide complex ion, Zn(OH)42-....
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...
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
Calculate the equilibrium molar concentration of free Zn2+(aq) in a solution that contains 0.021 mol of...
Calculate the equilibrium molar concentration of free Zn2+(aq) in a solution that contains 0.021 mol of Zn2+ per liter to which 0.498 mole of NH3 is added. Kf for Zn(NH3)42+ is 2.9x109. Show the equilibrium equation.
In the presence of excess OH-, the Al3+(aq) ion forms a hydroxide complex ion, Al(OH)4-. Calculate the concentration of...
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
This is all the information I was given to answer this question. To determine the formation...
This is all the information I was given to answer this
question.
To determine the formation constant, Kf, for [Zn(CN)4)2-(aq), 4.00 mL of 1.50 mol/L KCN(aq) and 1.00 mL of 0.0750 mol/L Zn(NO3)2(aq) were combined. The mixture was stirred for 10 minutes to ensure that equilibrium had been reached. A voltaic cell was constructed to determine the concentration of uncomplexed Zn2+ (aq) ions left in solution at the end of the complexation reaction. The [Zn(CN)4]- (aq), Zn2+ (aq), CN (aq)...
BACKGROUND: Synthesis of Potassium Iron (III) Oxalate Hydrate Salt The iron(II) ions from Fe(NH4)2(SO4)2•6H2O will be precipitated as iron(II) oxalate. Fe^2+(aq) + C2O4^2-(aq) --> FeC2O4 (s) The su...
BACKGROUND: Synthesis of Potassium Iron
(III) Oxalate Hydrate Salt
The iron(II) ions from
Fe(NH4)2(SO4)2•6H2O
will be precipitated as iron(II) oxalate.
Fe^2+(aq) + C2O4^2-(aq) --> FeC2O4 (s)
The supernatant liquid, containing the ammonium and sulfate
ions, as well as excess oxalate ions and oxalic acid will be
decanted and discarded. The solid will then be re-dissolved and the
iron(II) ions will be oxidized to iron(III) ions by reaction with
hydrogen peroxide.
2Fe^2+(aq) + H2O2(aq) --> 2Fe^3+(aq) +2 OH^ - (aq)
The...
Quest. 6 (20 pts). Nickel (II) ions forms a strong complex in the presence of ammonia (NH). Calculate the equilibrium concentrations of NP and [Ni(NHLP complex after 0.120 mol of Ni(NO3)2 has been dissolved in 1.00 L of 2.30 M NH, solution. The Kiof (Ni(NHSP is 2.0 x 10. N" (aq) +6 NHỊ (aq) # [NINH " (aq)
This is all the information I was given to answer this
question.
To determine the formation constant, Kf, for [Zn(CN)4)2-(aq), 4.00 mL of 1.50 mol/L KCN(aq) and 1.00 mL of 0.0750 mol/L Zn(NO3)2(aq) were combined. The mixture was stirred for 10 minutes to ensure that equilibrium had been reached. A voltaic cell was constructed to determine the concentration of uncomplexed Zn2+ (aq) ions left in solution at the end of the complexation reaction. The [Zn(CN)4]- (aq), Zn2+ (aq), CN (aq)...
BACKGROUND: Synthesis of Potassium Iron
(III) Oxalate Hydrate Salt
The iron(II) ions from
Fe(NH4)2(SO4)2•6H2O
will be precipitated as iron(II) oxalate.
Fe^2+(aq) + C2O4^2-(aq) --> FeC2O4 (s)
The supernatant liquid, containing the ammonium and sulfate
ions, as well as excess oxalate ions and oxalic acid will be
decanted and discarded. The solid will then be re-dissolved and the
iron(II) ions will be oxidized to iron(III) ions by reaction with
hydrogen peroxide.
2Fe^2+(aq) + H2O2(aq) --> 2Fe^3+(aq) +2 OH^ - (aq)
The...
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