A
When water is added to lead chloride, the solute sparingly dissolves with some undisclosed particles settling at the bottom of the beaker. Lead chloride is only sparingly soluble in water. When the dilute hydrochloric acid is added to the saturated solution of lead chloride, a white precipitate is formed. This is because of the common ion effect where the excess chloride ions from HCL cause lead chloride to precipitate.
Pb2+ aq + 2Cl-aq PbCl2 s
B
0.1M lead nitrate +0.1M sodium oxalate
Ionic Equations:
Pb2+ (aq) +C2O42- (aq) PbC2O4(s)
Na+ (aq) + NO3-(a) NaNO3 (aq)
Lead nitrate + sulphuric acid
Ionic equation
Pb2+ (aq) + SO42-(aq) PbSO4 (aq)
Some of the lead oxalate precipitate disappeared when nitric acid was added to the first test. This is due to the oxidation of lead oxalate by nitric acid. Nitric acid is a strong oxidizing agent; it thus oxidizes some of lead oxalate and that way the reducing the amount of the precipitate.
In the second test tube, lead sulphate does not dissolve in dilute acids (nitric acid).The amount of precipitate remains the same when dilute nitric acid is added .When lead react with sulfuric acid it forms lead II sulphate .A sulphate coating is formed on lead metal which prevents it from reacting with dilute acids. The sulphate layer obscures the oxidation of the compound by dilute acids.
The oxalate ions in the lead oxalate are easily displaced by the nitrate ions in the nitric acid. This explains why a substantial amount of the lead oxalate precipitate disappears. However, in the second scenario, the sulphate ions masks or coats the lead metal providing a masking effect which prevents it from reacting with the dilute nitric acid.
C1
When 1 drop of the barium chloride is added to the 10ml of 10% sodium sulphate solution, a white precipitate is formed. This is because of the formation of the insoluble barium sulphate as shown in the ionic equations below
SO42aq + Ba2+ aq + Cl-aq BaSO4s + Cl-aq
SO42aq + Ba2+ aq BaSO4
C2
When 10 ml of the 10% sodium sulphate solution is added to the 30% calcium chloride solution, a white precipitate is formed. This white precipitate is the insoluble calcium sulphate as shown in the equations below.
Ca2+(aq) + Cl- (aq) + Na+(aq) + SO4 2-(aq)=====>Na+(aq) + Cl- (aq) + CaSO4(s)
Ca2+(aq) + SO4 2-(aq) =====> CaSO4(s)
Calcium sulphate has a less solubility product numeral. This is because the calcium suphate is usually less soluble than barium suphate.
C3
Ca2+ (aq) +SO42- (aq) CaSO4 (aq)
There are less number of drops of calcium chloride required to produce a precipitation .This is due to increase in amount of sulphate (SO42-) ions available for precipitation.
D
Solubility Product Report sheet
Equation for the precipitation of silver acetate
Ag(s) + CH3COO- ---> AgCH3COO
Ksp expression for silver acetate
Ksp = [Ag+] [CH3COO-]
Volume of Potassium Thiocyanate Solution 50 (mL)
Molarity of Potassium Thiocyanate 0.02M
Temperature of filtrate 25°C
Trial 1 Trial 2 Trial 3
Final burette reading (ml) 12.42 22.63 32.74
Initial burette reading (ml) 2.12 12.42 22.63
Volume used (ml) 10.30 10.21 10.11
Moles of Thiocyanate added 0.00020 0.00020 0.0002
Calculation of moles
Molarity of thiocynate is 0.02M, but number of moles is given molarity x volume in liters
Thus in
Trial one 0.02 x 0.01030 = 0.00020 moles
Trial two 0.02 x 0.01021 = 0.00020 moles
Trial three 0.02 x 0.01011= 0.00020 moles
1. Amount of Ag+ present
Calculations
Here, the mole ratio of thiocynate to Ag+ is 1:1, Thus the moles of Ag+ is equal to the mole of thiocynate calculated above. To get the moles in 1000ml which is actually the molarity we multiply 1000ml with the number of moles and divide by 5.00ml
Trial 1 0.00020 x 1000/5 = 0.04
Trial 2 0.00020 x 1000/5 = 0.04
Trial 3 0.00020 x 1000/5 = 0.04
Trial 1 0.00020 moles of Ag+ in 5.00ml of solution 1
0.04 M (moles of Ag+ in 1000ml of solution 1 )
Trial 2 0.00020 moles of Ag+ in 5.00ml of solution 1
0.04 M (moles of Ag+ in 1000ml of solution 1)
Trial 3 0.00020 moles of Ag+ in 5.00ml of solution
0.04 M (moles of Ag+ in 1000ml of solution 1)
2. 9 x 10-3 moles of acetate ion in 70Ml of solution 1 (before precipitation).
Calculation
Amount of acetate ions in solution 1 is 30 ml .however the molarity of sodium acetate is 0.300M.The mole ratio of sodium and acetate ions in sodium acetate is 1 :1.Moles of acetate are thus 30/1000 x 0.300= 9 x 10-3 moles
3. Initial morality of acetate ions:
0.1286 Moles of acetate ion in 1000Ml of solution (before precipitation)
Therefore, original molarity of acetate ion in solution 1 is 0.1286 M.
Calculation
Moles of acetate ions in 70ml are 9 x 10-3 .Therefore in 1000ml of solution; moles present:
1000 x 9 x10-3 / 70 = 0.1286 moles. Hence molarity is 0.1286M
4. Moles of silver of ion initially present:
8 x 10-3 Moles of silver ion in 70 Ml of solution 1 (before precipitation)
Calculation
The mole ratio of silver and nitrate ions in silver nitrate is 1:1.The molarity of silver ions is thus 0.200 M. But the amount of silver ions in solution 1 is 40ml.Therefore a mole of silver ions in 70 ml is: 40/1000 x 0.200M = 8 x 10-3
5. Initial molarity of silver ions:
0.1143 Moles of silver ion in 1000Ml of solution 1 (before precipitation)
Therefore original molarity of silver ion in solution 1 is 0.1143 M.
Calculation
There are 8 x 10-3 moles of silver ions in the 70 ml (solution 1).Therefore, in a 1000ml of solution 1, the amount of silver ions in moles (molarity) is 1000 x 8 x 10-3 / 70 = 0.1143 M.
6.
Trial 1 Trial 2 Trial 3
- Original Silver ion concentration (from 5) 0.1143 0.1143 0.1143
- Molarity of silver ion remaining after precipitation
remaining after precipitation. (from 1) 0.04 0.04 0.04
- Molarity of silver ion precipitated (a-b) 0.0743 0.0743 0.0743
- Original acetate ion concentration 0.1286 0.1286 0.1286
- Molarity of acetate ion precipitated 0.04 0.04 0.04
- Acetate ion concentration remaining
remaining after precipitation (d-e) 0.0886 0.0866 0.0886
7. Ksp for silver acetate (b x f) 3.544 x 10-3 3.637 x 10-3 3.544 x 10-3
8. The Ksp for silver acetate is [(3.544 x 10-3) + (3.637 x 10-3 ) + (3.544 x 10-3)] / 3
= 3.575 x 10-3
Discussion Questions
Question 1
The molar mass of silver acetate is 166.91222, therefore, moles of silver acetate present in 1.02 g is 1.02/ 166.91222 .This yields 6.11 x 10-3 moles per 100ml of water. For a 1000ml of water and hence molar solubility: 6.11 x 10-3 x 10 which yields 6.11 x 10-2 mols/1000ml.
AgC2H3O2 Ag+ + C2H300-
Mole ratio= 1: 1
The Ksp expression is [Ag+] [C2H300-]
There is a 1: 1 ratio between [Ag+] and [C2H300-] ions. This means that when 6.11 x 10-2 moles/1000ml of AgC2H3O2 dissolves, it produces 6.11 x 10-2 moles/1000ml of [Ag+] and 6.11 x 10-2 moles/1000ml [C2H300-] ions. Inserting these values into the Ksp expression we get:
[6.11 x 10-2] [6.11 x 10-2] = 3.73321 x 10-3.
Question 2
Difference: 3.73321 x 10-3 - 3.575 x 10-3 =1.5821 x 10-4
Percentage difference 1.5821 x 10-4/3.73321 x 10-3 x 100=4.24%
The percentage difference percentage is 4.24%
Question 3
Usually, when an experiment is conducted to determine the Ksp value of a certain compound, the value obtained does not always agree with the literature value and this particular experiment is testament to this. The experimental value obtained may be larger than the literature value or may be smaller than it. In our experiment, the literature value is larger than the experimental value by 4.24%.
There are several reasons why this value may not always agree with the literature value
One reason is temperature variation. Differences in experimental temperature may yield different results. Solubility is usually regulated by temperature, with the effect being that salts readily dissolve at relatively higher temperatures. If the room temperature was higher than the theoretical one, the experimental Ksp will be lower as seen in our experiment.
The other reason would be an unsaturated silver acetate solution. If the silver acetate was not completely saturated or was just slightly saturated, the Ksp value obtained would be relatively lower.
The other reason would be the use of wet materials like the reagent flasks and filter paper. This wetness actually leads to some gravimetric alterations due to the inclusion of the moisture in various calculations. Consequently, the Ksp value obtained will not be the same as the theoretical one. For example, if the materials and reagents were not completely dry, the mass recorded would be higher and the Ksp calculated will also be higher.
Question 4
In quantitative experiments quantities of substances are analyzed for there to be consistence in results and to avoid gravimetric alterations or changes, the apparatus should be dry and devoid of any solution. In case there are liquids or solutions trapped in the filter paper, funnels or in the flask, the amounts of substances analyzed are altered. These alterations may occur as a result of the analytes reacting with the solutions or basically mixing up which alters the gravimetric amounts.
Predicates are in solid form and when it occurs that water or detergents used in cleaning the apparatus is still trapped in the flasks or the funnels, reaction may occur and thus altering the amount of solid available for reacting .Interferences from these solutions/liquids does not only lead to a change in gravimetric values, it may also alter the chemical composition and properties of the substances that are under analysis.
Conclusion
Works Cited
Hagenmuller, Paul. Inorganic Solid Fluorides: Chemistry and Physics. Orlando, Fla: Academic Press, 1985. Print.
Final Report on Effect of Concentration and Solubility of Organic Chemicals. S.l.: Bibliogov, 2013. Print.
"The Effect of Some Salts on the Solubility of Silver Acetate and of Silver Nitrate in Acetic Acid at 30°1 - Journal of the American Chemical Society (ACS Publications)." An Error Occurred Setting Your User Cookie. N.p., 12 Mar. 2012. Web. 9 June 2013.
