Can someone explain to me on how to find the rest?I\'m confused on how to get th
ID: 1032976 • Letter: C
Question
Can someone explain to me on how to find the rest?I'm confused on how to get the Ksp. I just need an example on how to do it. Thank you!!!!
A
?
?
Average
Initial vol of Silver Acetate (ml)
2.85 ml
11.80 ml
Final vol of Silver Acetate (ml)
11.80 ml
20.50 ml
Vol of Silver Acetate (ml)
8.95 ml
8.70 ml
Vol of 0.0500 M KCl (ml)
10.00 ml
10.00 ml
Moles of Cl- (mol)
0.0500 mol
0.0500 mol
Moles of Ag+ (mol)
0.0500 mol
0.0500 mol
Molarity of Ag+ [Ag+] (M)
Molarity Acetate [C?H?O?-] (M)
Ksp (Silver Acetate)
B (in 0.100 M KNO?)
?
?
Average
Initial vol of Silver Acetate (ml)
0 ml
8.00 ml
Final vol of Silver Acetate (ml)
8.00 ml
15.95 ml
Vol of Silver Acetate (ml)
8.00 ml
7.95 ml
Vol of 0.0500 M KCl (ml)
10.00 ml
10.00 ml
Moles of Cl- (mol)
0.0500 mol
0.0500 mol
Moles of Ag+ (mol)
0.0500 mol
0.0500 mol
Molarity of Ag+ [Ag+] (M)
Molarity Acetate [C?H?O?-] (M)
Ksp (Silver Acetate)
C (in 0.100 M AgNO?)
?
?
Average
Initial vol of Silver Acetate (ml)
19.00 ml
27.30 ml
Final vol of Silver Acetate (ml)
27.30 ml
35.35 ml
Vol of Silver Acetate (ml)
8.30 ml
8.05 ml
Vol of 0.0500 M KCl (ml)
25.00 ml
25.00 ml
Moles of Cl- (mol)
0.0500 mol
0.0500 mol
Moles of Ag+ (mol)
0.0500 mol
0.0500 mol
Molarity of Ag+ [Ag+] (M)
Ksp (Silver Acetate) (from Part A)
Molarity Acetate [C?H?O?-] (M)
D (in 0.100 M NaC?H?O?)
?
?
Average
Initial vol of Silver Acetate (ml)
0.9 ml
19.30 ml
Final vol of Silver Acetate (ml)
19.30 ml
37.65 ml
Vol of Silver Acetate (ml)
18.40 ml
18.35 ml
Vol of 0.0500 M KCl (ml)
10.00 ml
10.00 ml
Moles of Cl- (mol)
0.0500 mol
0.0500 mol
Moles of Ag+ (mol)
0.0500 mol
0.0500 mol
Molarity of Ag+ [Ag+] (M)
Ksp (Silver Acetate) (from Part A)
Molarity Acetate [C?H?O?-] (M)
A
?
?
Average
Initial vol of Silver Acetate (ml)
2.85 ml
11.80 ml
Final vol of Silver Acetate (ml)
11.80 ml
20.50 ml
Vol of Silver Acetate (ml)
8.95 ml
8.70 ml
Vol of 0.0500 M KCl (ml)
10.00 ml
10.00 ml
Moles of Cl- (mol)
0.0500 mol
0.0500 mol
Moles of Ag+ (mol)
0.0500 mol
0.0500 mol
Molarity of Ag+ [Ag+] (M)
Molarity Acetate [C?H?O?-] (M)
Ksp (Silver Acetate)
Explanation / Answer
A
Average
Initial Vol. of silver acetate (mL)
2.85
11.80
Final Vol. of silver acetate (mL)
11.80
20.50
Vol. of silver acetate (mL)
8.95
8.70
Vol. of 0.0500 M KCl (mL)
10.00
10.00
Moles of Cl- (mol)
(10.00 mL)*(1 L/1000)*(0.500 M) = 0.005
(10.00 mL)*(1 L/1000)*(0.500 M) = 0.005
Moles of Ag+ (mol)
Ag+ reacts with Cl- to form AgCl in a 1:1 molar ratio as
Ag+ (aq) + Cl- (aq) ------> AgCl (s)
Moles Ag+ = Moles Cl-. Therefore,
Moles Ag+ = moles Cl- = 0.005
Ag+ reacts with Cl- to form AgCl in a 1:1 molar ratio as
Ag+ (aq) + Cl- (aq) ------> AgCl (s)
Moles Ag+ = Moles Cl-. Therefore,
Moles Ag+ = moles Cl- = 0.005
Molarity of Ag+, i.e, [Ag+] (M)
(moles of Ag+)/(volume of solution in L) = (0.005 mole)/[(8.95 mL)*(1 L/1000 mL)] = 0.5586 mol/L
(moles of Ag+)/(volume of solution in L) = (0.005 mole)/[(8.70 mL)*(1 L/1000 mL)] = 0.5747 mol/L
Molarity of C2H3O2-, i.e, [C2H3O2-] (M)
Silver acetate AgC2H3O2 ionizes as
AgC2H3O2 (s) <====> Ag+ (aq) + C2H3O2- (aq)
Moles Ag+ = moles C2H3O2- = 0.005 mole.
Volume of the solution = 8.95 mL = (8.95 mL)*(1 L/1000 mL) = 0.00895 L.
Molarity of C2H3O2- = (0.005 mole)/(0.00895 L) = 0.5586 mol/L.
Molarity of C2H3O2- = (0.005 mole)/(0.00870 L) = 0.5747 mol/L.
Ksp
[Ag+][C2H3O2-] = (0.5586)*(0.5586) = 0.3120 ? 0.312
[Ag+][C2H3O2-] = (0.5747)*(0.5747) = 0.3302 ? 0.330
½*(0.312 + 0.330) = 0.321
A
Average
Initial Vol. of silver acetate (mL)
2.85
11.80
Final Vol. of silver acetate (mL)
11.80
20.50
Vol. of silver acetate (mL)
8.95
8.70
Vol. of 0.0500 M KCl (mL)
10.00
10.00
Moles of Cl- (mol)
(10.00 mL)*(1 L/1000)*(0.500 M) = 0.005
(10.00 mL)*(1 L/1000)*(0.500 M) = 0.005
Moles of Ag+ (mol)
Ag+ reacts with Cl- to form AgCl in a 1:1 molar ratio as
Ag+ (aq) + Cl- (aq) ------> AgCl (s)
Moles Ag+ = Moles Cl-. Therefore,
Moles Ag+ = moles Cl- = 0.005
Ag+ reacts with Cl- to form AgCl in a 1:1 molar ratio as
Ag+ (aq) + Cl- (aq) ------> AgCl (s)
Moles Ag+ = Moles Cl-. Therefore,
Moles Ag+ = moles Cl- = 0.005
Molarity of Ag+, i.e, [Ag+] (M)
(moles of Ag+)/(volume of solution in L) = (0.005 mole)/[(8.95 mL)*(1 L/1000 mL)] = 0.5586 mol/L
(moles of Ag+)/(volume of solution in L) = (0.005 mole)/[(8.70 mL)*(1 L/1000 mL)] = 0.5747 mol/L
Molarity of C2H3O2-, i.e, [C2H3O2-] (M)
Silver acetate AgC2H3O2 ionizes as
AgC2H3O2 (s) <====> Ag+ (aq) + C2H3O2- (aq)
Moles Ag+ = moles C2H3O2- = 0.005 mole.
Volume of the solution = 8.95 mL = (8.95 mL)*(1 L/1000 mL) = 0.00895 L.
Molarity of C2H3O2- = (0.005 mole)/(0.00895 L) = 0.5586 mol/L.
Molarity of C2H3O2- = (0.005 mole)/(0.00870 L) = 0.5747 mol/L.
Ksp
[Ag+][C2H3O2-] = (0.5586)*(0.5586) = 0.3120 ? 0.312
[Ag+][C2H3O2-] = (0.5747)*(0.5747) = 0.3302 ? 0.330
½*(0.312 + 0.330) = 0.321