The formation constant for triethylenetetraamine - silver is5.0x10^7. If the complex formed is 1:1 Ag + Trien,calculate the concentration of Ag + in equilibrium when10.00 mL.0250 M of trien solution is added to 20.00mL ofAg+ .0120M. (Verify limiting reactant to help solve).Please show a step by step solving to better myunderstanding of the process

Given: VTETA=10.0mL;[TETA]=0.0250M;[Ag+]=0.0125M;VAg+=20.0mLKf(Ag(TETA)+)=5.0x107;[Ag+]=???MStep…

Answered: The formation constant for…

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter16: Solubility And Precipitation Equilibria

Section: Chapter Questions

Problem 47P

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The molecular complex between silver ion and cyanide ion forms by the reactionAg+(aq) + 2 CN-(aq) ⇌ Ag(CN)2-(aq) Kf = 1.0 x 10^21A solution initially has [Ag+] = 2.0 x 104 M, and [CN-] = 0.0400 M. What will be the concentration ofAg+ ion at equilibrium?
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Which of the following is the correct Kf expression for the formation of the complex ion tetraiodoplumbate(II) ion, [PbI4]2–? Pb2+ (aq) + 4 I1- (aq) < === > [PbI4]2– (aq) Kf = [PbI4]2– / {[Pb2+] [4 I1-]} Kf = {[Pb2+] 4[I1-]} / [PbI4]2– Kf = {[Pb2+] [I1-]4} / [PbI4]2– Kf = [PbI4]2– / {[Pb2+] [I1-]4}
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In the presence of aqueous cyanide, cadmium(II) ions will form the complex Cd(CN)42- . Determine the molar concentration of free( uncomplexed) cadmium(II) ion in solution when 47.3g of Cd(NO3)2 is dissolved in exactly one liter of 2.0M sodium cyanide Kf of Cd(CN)42- at 25 degrees C is 7.2x1016
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