Moles SCN- used in titration (mol) Moles of Agt in 20.0 mL titrated (mol) [Ag*lequil (M) X= [Ag*Jinit - [Ag*lequil = Change (M) solve for flask d e and f
Moles SCN- used in titration (mol) Moles of Agt in 20.0 mL titrated (mol) [Ag*lequil (M) X= [Ag*Jinit - [Ag*lequil = Change (M) solve for flask d e and f
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
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Question
- Moles SCN- used in titration (mol)
Moles of Agt in 20.0 mL titrated (mol) - [Ag*lequil (M)
X= [Ag*Jinit - [Ag*lequil = Change (M)
solve for flask d e and f
![C. Calculations
1. Assume that the total volume of each saturated solution is 55.00 mL.
2. When you prepare this equilibrium by mixing solutions, you are approaching this equilibrium
from the right. Use an 'ICE Table' approach to getting the equilibrium concentrations
[Ag leq and [C₂H302 leq in each of your six flasks.
AgC₂H3O2 (s)
Initial
Change
Equilibrium
3. The Initial [Aglinit and [C₂H3O₂ linit concentrations can be calculated from the molarities
and volumes of solutions used. The initial concentration refers to the system after mixing but
before precipitation. You will need to refer to the tables on pages 2 and 3 to get this
information.
example a: [Ag+]init
Ag+ (aq) + C₂H3O₂ (aq)
[C₂H3O2 linit
[Ag+] init
-X
([Ag+]init - X)
total moles of Ag+ put into flask a
total liters solution in flask a.
in sample a.
4. The Change referred to in the' ICE table' occurs when precipitate forms after mixing. The
equilibrium shifts to the left, reducing the silver and the acetate ion concentrations by X
mol/L.
=
=
0.0350 L solutions x 0.250 mol Ag/L solution
0.0550 L
= 0.159 M
-X
([C₂H3O2-linit - X)
5. In each of your samples, you use your titration volume, V mL, to calculate the total moles of
Ag+ in the 20.0 mL sample of saturated solution you titrated by doing a stoichiometry
calculation using the chemical equation for the titration reaction.
Ag+ (aq) + SCN- (aq) -> AgSCN(s)
?mol
V mL
0.0500 M
V mL KSCN solution x 0.0500 mol KSCN x
1000 mL KSCN sol
6. Calculate [Ag+] at equilibrium
[Ag*lequil
1 mol SCN- x
1 mol KSCN
1 mol Ag+
1 mol SCN-
(5.00 x 10-5 x V) moles Ag+ in 20.0 mL
(5.00 x 10-5 x V moles Ag+) = (2.5 x 10-3 x V) mol/L
0.0200 L solution](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9b0e7686-0c0d-4eba-ac81-54e736b47f2f%2Fec03efb2-85b5-4faa-b497-c8f09ab5f26b%2Fgdjyqjv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:C. Calculations
1. Assume that the total volume of each saturated solution is 55.00 mL.
2. When you prepare this equilibrium by mixing solutions, you are approaching this equilibrium
from the right. Use an 'ICE Table' approach to getting the equilibrium concentrations
[Ag leq and [C₂H302 leq in each of your six flasks.
AgC₂H3O2 (s)
Initial
Change
Equilibrium
3. The Initial [Aglinit and [C₂H3O₂ linit concentrations can be calculated from the molarities
and volumes of solutions used. The initial concentration refers to the system after mixing but
before precipitation. You will need to refer to the tables on pages 2 and 3 to get this
information.
example a: [Ag+]init
Ag+ (aq) + C₂H3O₂ (aq)
[C₂H3O2 linit
[Ag+] init
-X
([Ag+]init - X)
total moles of Ag+ put into flask a
total liters solution in flask a.
in sample a.
4. The Change referred to in the' ICE table' occurs when precipitate forms after mixing. The
equilibrium shifts to the left, reducing the silver and the acetate ion concentrations by X
mol/L.
=
=
0.0350 L solutions x 0.250 mol Ag/L solution
0.0550 L
= 0.159 M
-X
([C₂H3O2-linit - X)
5. In each of your samples, you use your titration volume, V mL, to calculate the total moles of
Ag+ in the 20.0 mL sample of saturated solution you titrated by doing a stoichiometry
calculation using the chemical equation for the titration reaction.
Ag+ (aq) + SCN- (aq) -> AgSCN(s)
?mol
V mL
0.0500 M
V mL KSCN solution x 0.0500 mol KSCN x
1000 mL KSCN sol
6. Calculate [Ag+] at equilibrium
[Ag*lequil
1 mol SCN- x
1 mol KSCN
1 mol Ag+
1 mol SCN-
(5.00 x 10-5 x V) moles Ag+ in 20.0 mL
(5.00 x 10-5 x V moles Ag+) = (2.5 x 10-3 x V) mol/L
0.0200 L solution
Transcribed Image Text:Solution 0.300 M NaC₂H3O₂
d
25.0 mL
25.0 mL
25.0 mL
e
f
0.250 M AgNO3
30.0 mL
30.0 mL
30.0 mL
Temp. °C
ice bath
room temp
50 - 60°C
AgC₂H3O₂ (s) <=> Ag¹(aq) + C₂H₂O₂ (aq)
<--- Equilibrium approached from right.
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