Can someone show me how to get the rest? I just need an example on how to finish
ID: 1032158 • Letter: C
Question
Can someone show me how to get the rest? I just need an example on how to finish 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
Hello,
The question you are trying to solve uses the Mole Concept & the concept of Solubility Product. You might be aware that, an acid and base react to form salt and water. Salts are classified into two categories on the basis of their solubility in water:
a) Soluble Salts – these salts are completely soluble in water, which suggests that all molecules in a given sample would completely dissociate into the constituent ions, giving rise to a clear solution For example: NaCl, KCl, KNO3, AgNO3, CH3COONa etc
b) Sparingly Soluble Salts - these salts aren’t completely soluble in water, which suggests that a given sample of these salts wouldn’t completely dissociate in water. Some part would, and a major part would remain insoluble resulting in a solid insoluble part in the solution or a turbid solution. For example: AgCl, CH3COOAg, BaSO4 etc
Now if we represent a salt as A-B, it’d undergo dissolution to produce A+ & B- ions. If we treat this dissolution process as an equilibrium then the equilibrium equation can be represented as
A-B(solid) + aq. = A+(aq) + B-(aq)
For this equilibrium if the equilibrium constant is expressed as K, then it’d be related to the concentration of species as:
K = [A+(aq)][B-(aq)]/[A-B(undissociated)]
Now if we consider
This equilibrium constant (K), when multiplied with the concentration of the undissociated salt, gives a new constant Ksp, which is known as the solubility product. As you can see it holds relevance only for a sparingly soluble salt, since concentration of its undissociated part would be very large and fairly constant at different dilutions.
# Solubility product is calculated at the stage when a saturated solution is obtained as at only that stage, an equilibrium would result.
# We can also express the value of Ksp in terms of the solubility of the respective ions derived from the dissociation of a sparingly soluble salt in the following manner:
Let degree of ionization of weak electrolyte AmBn be ‘?’ & solubility in moles/litres be ‘s’
AmBn = mAn+ + nBm-
s 0 0
teq s-s? ms? ns?
Ksp = [An+]m[Bm-]n = [ms? ]m[ns?]n
Ksp = mmnn (s?)m+n
Now in your question we’ll have to calculate the solubility product, which can be done quite easily by identifying
1. Moles of reactants
2. Identifying the limiting reagent
3. Identifying the number of moles product
4. Identifying the number of ions while common ion effect takes place.
Part A
For sample 1
Volume of CH3COOAg = 8.95 ml
Molarity of KCl = 0.05 M
Volume of KCl = 10 ml
Moles of Cl- obtained from KCl = 0.05 moles
Since in this reaction KCl would react with Ag+ from CH3COOAg, in quantitative amounts so the moles of KCl used is equal to the moles of Ag+ i.e. 0.05 M of KCl x 10 ml of KCl = 0.5 mmol of KCl = 0.5 mmol of Ag+
Since volume of CH3COOAg used is 8.95 ml, so molarity of [Ag+] = 0.5 mmol/8.95 ml = 0.0558 M
For Sample 2
Volume of CH3COOAg = 8.70 ml
Molarity of KCl = 0.05 M
Volume of KCl = 10 ml
Moles of Cl- obtained from KCl = 0.05 moles
Since in this reaction KCl would react with Ag+ from CH3COOAg, in quantitative amounts so the moles of KCl used is equal to the moles of Ag+ i.e. 0.05 M of KCl x 10 ml of KCl = 0.5 mmol of KCl = 0.5 mmol of Ag+
Since volume of CH3COOAg used is 8.95 ml, so molarity of [Ag+] = 0.5 mmol/8.70 ml = 0.0574 M
Using data from both Sample 1 & Sample 2 - Mean molarity turns out to be = (0.0574 + 0.0558)/2 = 0.0566 M
Since CH3COOAg dissociates in the following manner:
CH3COOAg = CH3COO- + Ag+
So the molar concentration of Ag+ has to be equal to the molar concentration of CH3COO-. Therefor the Ksp of CH3COOAg can be calculated as:
Ksp of CH3COOAg = [CH3COO-][Ag+] = [0.0566][0.0566] = 0.0032 M2 = 3.2 x 10-3
Part D (in CH3COONa)
For sample 1
Volume of CH3COOAg = 18.35 ml
Molarity of KCl = 0.05 M
Volume of KCl = 10 ml
Moles of Cl- obtained from KCl = 0.05 moles
Since in this reaction KCl would react with Ag+ from CH3COOAg, in quantitative amounts so the moles of KCl used is equal to the moles of Ag+ i.e. 0.05 M of KCl x 10 ml of KCl = 0.5 mmol of KCl = 0.5 mmol of Ag+
Since volume of CH3COOAg used is 8.95 ml, so molarity of [Ag+] = 0.5 mmol/18.40 ml = 0.0271 M
For Sample 2
Volume of CH3COOAg = 18.40 ml
Molarity of KCl = 0.05 M
Volume of KCl = 10 ml
Moles of Cl- obtained from KCl = 0.05 moles
Since in this reaction KCl would react with Ag+ from CH3COOAg, in quantitative amounts so the moles of KCl used is equal to the moles of Ag+ i.e. 0.05 M of KCl x 10 ml of KCl = 0.5 mmol of KCl = 0.5 mmol of Ag+
Since volume of CH3COOAg used is 8.95 ml, so molarity of [Ag+] = 0.5 mmol/18.35 ml = 0.0272 M
Using data from both Sample 1 & Sample 2 - Mean molarity turns out to be = (0.0271 + 0.0272)/2 = 0.02717 M
Since CH3COOAg dissociates in the following manner:
CH3COOAg = CH3COO- + Ag+
So the molar concentration of Ag+ has to be equal to the molar concentration of CH3COO-. Therefor the Ksp of CH3COOAg can be calculated as:
Ksp of CH3COOAg = [CH3COO-][Ag+] = [0.02717][0.02717] = 0.0007 M2 = 7 x 10-4 M2
I’ve solved two parts for you. You can repeat the same procedure for Part B & Part C which uses KNO3 & AgNO3. Most parts of the special characters in the question are not getting displayed. I’ve intuitively guessed the nature of compounds you are using for Part B, C & D.
Hope this helps. I request you to take some time, to rate the answer & to drop in your valuable feedback.
Please do post your queries pertaining to the solution provided (if any). hanks