Only need help with d-k. A solution prepared by mixing 60.0 mL of 0.120 M AgNO3
ID: 511589 • Letter: O
Question
Only need help with d-k.
A solution prepared by mixing 60.0 mL of 0.120 M AgNO3 and 60.0 mL of 0.120 MTINO3 was titrated with 0.240 M NaBr in a cell containing a silver indicator electrode and a reference electrode of constant potential 0.175 V. The reference electrode is attached to the positive terminal of the potentiometer, and the silver electrode is attached to the negative terminal. The solubility constant of TIBr is KS 3.6 x 10 and the solubility constant of AgBr is K 5.0 x 10 (a) Which precipitate forms firs O TIBr AgBr (b) Which of the following expressions shows how the cell potential, E, depends on CAg (I E 0.799-0.059 16loglAg J 0.175 O II (II) E-0.175- 0.799-0.059 16logl Ag O IV (II) E 0.799 0.05916log 0.175 Ag Scroll down to v) E-0.175- 0.799- 0.05916log T view the rest of the question. AgExplanation / Answer
As mentioned by you, help is required in Q (d-k)
Cell potential depends on the number of free Ag+ ions in the solution.
The AgNO3 solution is 0.12 M or has 0.12 Moles per liter.
Hence 60 mL of this solution will have 0.12 x (60/1000) = 0.0072 Moles
The reaction of AgNO3 with NaBr is a 1:1 stoichiometry to form AgBr & NaNO3
0.24 M NaBr has 0.24 moles per liter which would imply 0.24 x 10-3 moles per mL
d) 1 mL = 0.24 x 10-3 moles of NaBr will react with same moles of AgNO3
Remaing Ag+ ions: 7.2 x 10-3 - 0.24 x 10-3 = 6.96 x 10-3 which are present in 121 mL of solution
Hence, conc of Ag+ ions = 1000 x (6.96 x 10-3)/ 121 = 0.0575 M
Plug this in the equation in part b, and you will get your cell potential.
e) 16.3 mL = 16.3 x 0.24 x 10-3 = 3.912 x 10-3 moles of NaBr will react with same moles of AgNO3
Remaing Ag+ ions: 7.2 x 10-3 - 3.912 x 10-3 = 3.288 x 10-3 which are present in 136.3 mL of solution
Hence, conc of Ag+ ions = 1000 x (3.288 x 10-3)/ 136.3 = 0.0241 M
Plug this in the equation in part b, and you will get your cell potential.
f) 29 mL = 29 x 0.24 x 10-3 = 6.96 x 10-3 moles of NaBr will react with same moles of AgNO3
Remaing Ag+ ions: 7.2 x 10-3 - 6.96 x 10-3 = 0.24 x 10-3 which are present in 149 mL of solution
Hence, conc of Ag+ ions = 1000 x (0.24 x 10-3)/ 149 = 0.001611 M
Plug this in the equation in part b, and you will get your cell potential.
g) 29.9 mL = 29.9 x 0.24 x 10-3 = 7.176 x 10-3 moles of NaBr will react with same moles of AgNO3
Remaing Ag+ ions: 7.2 x 10-3 - 7.176 x 10-3 = 0.024 x 10-3 which are present in 149.9 mL of solution
Hence, conc of Ag+ ions = 1000 x (0.024 x 10-3)/ 149.9 = 1.601 x 10-4 M
Plug this in the equation in part b, and you will get your cell potential.
h) Now we have crossed the first equivalnce point. The Ag+ ions have reacted with the NaBr till 30 mL of addition. The AgBr salt has now precipitated and further addition of NaBr will result in precipitation of TlBr.
30.3 mL addition. 30 mL of NaBr added will precipitate as the silver salt. 0.3 mL in excess reacts with the Tl+ ion to form TiBr.
At the first equivalence point, the Br- ions added = Ag+ ions present. Hence Ksp = [Ag+][Br-] = [Ag+]2
Beyond this, till second equivalence point, the Br- ions added will react with Tl+ ions. We can assume since Ksp of AgBr << TlBr, the [Ag+] ions in solution will be same.
[Ag+] = Ksp0.5 = 7.071 x 10-7
Plug this in equation of cell potential as in "b" and we can calculate the same.
This can be used for cases "i" and "j" as well.
k) We have now crossed the second equivalence point.
The added NaBr solution will result in excess free Br- ions in the solution.
We know Ksp = [Ag+][Br-]. We know the free Br- ion concentration as (62.6 - 60) mL of 0.24 M NaBr
= 0.24 moles x 2.6/1000 = 6.24 x 10-4 moles in 182.6 mL
5 x 10-13 = [Ag+][6.24 x 10-4]
[Ag+] = 8.013 x 10-10
plug this in equation in part b and you will get the cell potential.
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