E. (5 pts sketch the general shape of the pTI vs.volume titration curve the titr

Question

E. (5 pts sketch the general shape of the pTI vs.volume titration curve the titration of 25.0 3+. for mL of a 0.0100 M Ti (aq) solution with o 0100 M aqueous EDTA at pH 250. both curve of has the shape flattened letter to as a sigmoidal curve label axes with the identity of of a a four items: the initial pTI: the the variable and show approximate ic scale for these post- equivalence point point plateau, equivalence int, and the plateau. Titration plot o 1 S o 2 is 22 2s 27 30 3 S 2.17 p as 308 602

Explanation / Answer

pka of EDTA=2.0,2.7,6.2,10.31

at pH=2.5

So,only one acetate of EDTA will be deprotonated ,with pKa>pH ie 2.0

Tl3+ + 3e------------>Tl

3Y-3e -3H+-------------->3Y-      where Y-=(EDTA)

----------------------------------------------------------------

Tl3+ +3Y ------------>Tl +3Y- + 3H+ overall rxn

1 mole Tl3+ reacts with 3 moles Y

F) moles of EDTA=0.01 mol/L*12.5*10^-3 L=12.5*10^-5 moles

moles of Tl3+ =0.01 mol/L *25*10^-3 L=25*10^-5 moles

moles of Tl3+ reacted= 3* moles of EDTA=3* 12.5*10^-5 moles=37.5*10^-5 moles

EDTA is in excess,so amount of Tl3+ actually reacted=1/3 *12.5*10^-5 moles=4.2*10^-5 moles

[Tl]formed=[Tl3+]=4.2*10^-5 moles

[Tl]=4.2*10^-5 moles/(25+12.5)*10^-3 L=0.112*10^-2 mol/L

pTl=-log[Tl]=-log(0.11*10^-2)=2.95

G) At equivalence point ,

moles of Tl3+= as all the Tl3+ will be used up.so[Tl]=[Tl3+] initial=25*10^-5 moles

[Tl]=25*10^-5 moles/(25+75)*10^-3 L=25*10^-4 mol/L

pTl=-log[Tl]=-log (25*10^-4 moles)=2.6

H)moles of EDTA=37.5*10^-3 L*0.01 mol/L=37.5*10^-5 moles

moles of Tl3+ reacted=1/3*37.5*10^-5 moles=12.5*10^-5 moles=[Tl] formed

[Tl]=12.5*10^-5 moles/(25+37.5ml)*10^-3 L/ml=0.2*10^-2 mol/L

pTl=-log(0.2*10^-2 mol/L)=2.7

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