In 'Synthesis of an Iron Complex Salt', a student added 95% ethyl alcohol to a solution containing iron and oxalate ions. This was done to titrate the solution until its endpoint create a metal-chelate complex

In 'Synthesis of an Iron Complex Salt', a student added 95% ethyl alcohol to a solution containing iron and oxalate ions. This was done to titrate the solution until its endpoint create a metal-chelate complex

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Determine the equilibrium concentration of nickel(II) ion in a solution that contains 2.3 M ammonia and 0.040 M nickel(II) nitrate. The formation constant for the complex [Ni(NH,) ] - 5.5×10® is 3.65×10-37 M 1.26×10-12 M 1.32×1054 M 3.65×10-10 M
Solid KSCN was added to a 2.25 M Co+ solution so that it was also initially 2.25 M SCN. These ions then reacted to give the complex ion Co(SCN)+, whose formation constant was 1.0 x 10°. What is the concentration of Co²+ (ag) at equilibrium? Be sure to check any simplifying assumption you make. M
Explain why addition of NH3 or HNO3 to a saturated solution of Cu(OH)2 in contact with solid Cu(OH), increases the solubility of the solid. O NH3 forms a complex ion with OH-, reducing its concentration and allowing more OH- ions to dissolve from the solid phase. HNO3 reacts with Cu2", thereby reducing the Cu2 concentration and allowing more solid to dissolve O NH3 forms a complex ion with Cu2, increasing the Cu2 concentration. HNO3 reacts with OH-, thereby increasing the OH- concentration and allowing more solid to dissolve O NH3 forms a complex ion with Cu2, reducing the Cu2 concentration and allowing more Cu2 ions to dissolve from the solid phase. HNO3 reacts with OH-, thereby reducing the OH- concentration and allowing more solid to dissolve
Using chemical equations, explain the effect of added OH- to the iron(III) thiocyanate complex formation equilibrium.
How would the addition of AgNO3 affect equilibrium system (D)? 2+ 2- [Co(H20)6] (ag) + 4 Cl (ag) A.The solution would become bluer since the silver would precipitate with the chloride ions shifting the = [Co(H2O)2CI4] (ag) + 4 H2O() equilibrium to the left. B. The solution would become pinker since the silver would precipitate with the chloride ions shifting the equilibrium to the left. C. The solution would become pinker since the silver would precipitate with the chloride ions shifting the equilibrium to the right. D. The solution would become bluer since the silver would precipitate with the chloride ions shifting the equilibrium to the right. I D Focus W
The formation constant of a silver cyano complex is 5.6 x 1018 Ag+ + 2 CN Ag(CN)," If a certain solution have the equilibrium concentrations of 5.0 x 10-3 M and 0.20 M for Ag+ and CN", respectively, then what is the equilibrium concentration for the silver cyano complex? O1.1 x 1015 O5.6 x 1015 02.8 x 1016 02.4 x 109
Take into consideration metal Z(III) ion which is able to form octahedral complexes with elther fluoride or cyanide ions, What can be said about the colors of the complexes formed when one complex ion absorbs at 650 nm while the other at 430 nm?
After completing the experiment, a student finds that the equilibrium concentrations of the reactants and product are as follows. Calculate the equilibrium constant for the complex ion formation. [SCN¯]eg= 6.45x10-4 M %3D [Fe3+]eg = 8.42x10-4 M leq [FENCS2+]eg = 1.50x10-4 M %3D
You are given a table of [CN-] standard solutions of different concentration. A 25 mL aliquot of each standard solution was added with Fe(NO3)3 to form a colored complex[Fe(CN)6]3-(aq). Each standard solution was then diluted to a total volume of50.0 mL. [CN–] prior to dilution, M [CN-] final M 0 0 0.04 0.1 0.16 0.24 0.32 0.4 What are the new concentrations for the standard solutions? Please show solutions.
Formation constant and Gibbs free energy For the complex formation reaction where Mis any metal and L is any ligand, M+ + 2L = [ML2]* Kr is the formation constant for the complex at equilibrium and AG is the Gibbs free energy change for the reaction at equilibrium. The formation constant (K) can be related to the Gibbs free energy change (AG°) for a reaction by the equation AG° = RT In Kr where T is the temperature in kelvins and R is the gas constant equal to 8.314 J K. mol. The complex formation is generally a spontaneous process; thus, the Gibbs free energy change is negative. Part B Calculate the formation constant for the formation of [Cu(en) (H2O)4 from (Cu(H20)6]+, given that AG° is -60.1kJ - mol at 298 K.
Ni2* forms a complex with ammonia (NH3) which has the formula [Ni(NH:)6]?*. Calculate the [Ni2*], [NH3], and [Ni(NH:)6"]´at equilibrium, when 0.300 M of [Ni²*] and 1.200 M of [NH3] are present in the solution initially. [Kr for [Ni(NH3)6]?* is 5.5x10$] Ni forms a complex with ammonia (NH) which has the formula [Ni(NH) J+ Calculate the [Ni). [NH), and [Ni(NH.)"] at equilibrium, when 0.300 M of [Ni ] and 1.200 M of [NH] are present in the solution initially. [Kr for [Ni(NH.)F is 5.5x10 ]
Please use the values in the resources listed below instead of the textbook values. Sometimes equilibria for complex ions are described in terms of dissociation constants, Ka. For the complex ion Cu(CN)4 3- the dissociation reaction is as follows. [Cu*]• [CN 14 Kd [Cu(CN)43-] Cu(CN)4 3- = Cu+ + 4 CN- = 5.0X10° n-31 = Calculate the value of the formation constant, Kf, for Cu(CN)4°¯. 4.0