Exp8-preparation And Studies Of Potassium Tris(oxalato)aluminate(iii) Trihydrate {k3[al(c2o4)3].3h2o}.docx

UDEC1134 Chemistry Laboratory I Bachelor Of Science (HONS) Chemistry

Name

Pek Chen Yee

Student ID

1704650

Members name

Toh Ka Yie Thong Siew Hui P2

Practical Group

Practical Subgroup Group 1 No. of experiment

Experiment 8

Title of experiment Preparation and Studies of Potassium Tris(Oxalato)Aluminate(III) Trihydrate {K3[AL(C2O4)3].3H2O}

Date of experiment 28/1/2019 Date of submission 11/2/2019 Name of lecturer

Ms. Chang Chew Cheen

Title Preparation and studies of potassium tris(oxalato)aluminate(III) trihydrate {K3[Al(C2O4)3].3H2O} Objective To prepare Potassium tris(Oxalato)aluminate(III)trihydrate {K3 [Al(C2O4 ) 3 ].3H2O} and study its reactions. Apparatus Beaker Measuring cylinder Büchner funnel Büchner flask Hot plate Glass rod Filter paper Ice Platinum wire Bunsen burner Weighting boat Dropper Materials 0.005 mole Al2(SO4)3.18H2O 5M NaOH 0.008 mole oxalic acid dihydrate 0.008 mole potassium oxalate monohydrate Ethanol Dilute sodium hydroxide solution Dilute HCl Dilute H2SO4 Dilute solution of KMnO4

Results Mass of Al2(SO4)3.18H2O

3.12g

Mass of oxalic acid dihydrate

1.08g

Mass of potassium oxalate monohydrate

1.47g

Mass of weighting board

0.7780g

Mass of product + weighting board

2.1785g

Mass of product

1.4005g

Test for potassium Colour of the flame: The purplish pink colour flame was formed.

Test for aluminium Before warm the solution and add NaOH No precipitate was formed solution Before warm the solution and add NaOH No precipitate was formed solution

Test for oxalate With adding H2SO4

The purple solution turn to colourless quickly

Without adding H2SO4

The purple solution turn to colourless slower than the solution with adding H2SO4

Calculation Al2(SO4)3.18H2O (s) + 6NaOH (aq) → 2Al(OH)3 (s) + 3Na2SO4 (aq) + 18H2O (l) 1 mol of Al2(SO4)3.18H2O produce 2 mol of Al(OH)3 Hence, 0.005 mol of Al2(SO4)3.18H2O produce 0.010 mol of Al(OH)3

2Al(OH)3 (s) + 3K2C2O4 (aq) + 3H2C2O4 (aq) → 2K3[Al (C2O4)3].3H2O(l) 2 mol of Al(OH)3 produce 2mol of K3[Al (C2O4)3].3H2O Hence, 0.010 mol of Al(OH)3 produce 0.010 mol of K3[Al (C2O4)3].3H2O

The product, K3[Al (C2O4)3].3H2O Molar mass of K3[Al (C2O4)3].3H2O =3(39.098) + 26.982 + 3[2(12.011) + 4(15.999)] + 3[2(1.0079) + 15.999] =462.3744 g/mol

Theoretical yield of K3[Al (C2O4)3].3H2O = 0.010 mol×462.3744g/mol = 4.6237 g

Questions 1) Write an ionic equation for the formation of Al(OH)3 using Al2(SO4)3.18H2O and NaOH. The equation: Al2(SO4)3 (aq) + 6NaOH (aq) → 2Al(OH)3 (s) + 3Na2SO4 (aq) Ionic equation: 2Al3+(aq)+3SO42-(aq)+6Na+(aq)+6OH-(aq) →2Al(OH)3 (s)+6Na+(aq)+ 3SO42-(aq) Net ionic equation: 2Al3+(aq) + 6OH- (aq) → 2Al(OH)3 (s)

2) The formation of the tris(oxalate)aluminate(III) complex anion is carried out by reacting Al(OH)3 with the exact number of moles of the oxalate ion and oxalic acid. The number of moles of each reactant is given in step (3) of the preparation section. Using this information, balance the equation for the reaction: 2Al(OH)3 + 3(C2O4 2- ) + 3H2C2O4.2H2O => 2[Al(C2O4)3]3-.3H2O + 12H2O

3) Aluminum is present as a free cation when Al2(SO4)3.18H2O is dissolved in water. i) Which ions are present when K3[Al(C2O4)3].3H2O is dissolved in water? The function of acid used in carrying out test (ii) is to release Al3+ from the complex ion, [A l(C2O4)3]3- . The [Al3+(H2O)6] ion written in simple form as Al3+ is then identified as described in test (ii). Write equations for the reaction of the acid with the product and for the reaction which indicates the presence of aluminum (III) in the product. i) ii)

K+ and [Al(C2O4)3]3- are the ions present in the water. [Al(C2O4)3]3- (aq) + HCl (aq) → AlCl3 (aq) + 3H+ (aq) + 6CO2(g)

4) Aluminum (III) hydroxide is an amphoteric substance i.e., it reacts both with acids and alkalis to form the aluminum salt of the acid and AlO2 - anion respectively. Write equations for the reactions of Al(OH)3 with HCl and NaOH to indicate the amphoteric nature of Al(OH)3. Reaction Al(OH)3 with HCl: Al(OH)3 + 3HCl → AlCl3 + 3H2O (l) Reaction Al(OH)3 with NaOH: Al(OH)3 + NaOH →Na[Al(OH)4]

5) The oxalate ion is oxidised by the permanganate ion in the acid solution. Follow the following steps to derive the two half ionic equations for this reaction: Oxidation of oxalate ion to carbon dioxide (i) Write C2O4 2- on the left and CO2 on the right of =>. (ii) Balance the carbon and oxygen atoms. (iii) Balance the charge by writing electrons (e-) on the right hand-side. Why is it called an oxidation reaction? Reduction of MnO4 - to Mn2+ in acid solution

Write MnO4- on the left and Mn2+ on the right of =>. Balance oxygen by writing H2O on the right hand-side. Balance hydrogen by writing H+ on the left hand-side. Balance the charge by writing electrons (e-) on the left hand-side.

(i) (ii) (iii) (iv)

Multiply with the appropriate factor to obtain the same number of electrons in each half equation. Add the two half-equations to obtain the overall equation for the oxidation reduction reaction. Why is it called reduction reaction? What evidence did you see for this redox reaction in test (iii) to indicate the presence of oxalate in your product? Equation 1: C2O4 2- → 2CO2 + 2eIn this reaction, C2O4 2- increase number of oxidation state and loss 2eEquation 2: MnO4- + 8H+ + 5e- → Mn2++ 4H2O Reduction is shown by the permanganate ion as it obtained the electrons. From the equation below, we can see that Mn decrease from +7 to +2. +7

+2

MnO4- → Mn2+ From the half equation C2O4 2- → 2CO2 + 2e-

] ×5

MnO4- + 8H+ + 5e- → Mn2++ 4H2O

] ×2

Adding equations: 2MnO4- + 16H+ + 10e- → 2Mn2++ 8H2O 5C2O4 2- → 10CO2 + 10e________________________________________________________________ Overall equation: 2MnO4- + 5C2O4 2- + 16H+ → 2Mn2+ + 10CO2 + 8H2O

6) Calculate the % yield of the product. 𝑎𝑐𝑡𝑢𝑎𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡

Given that percentage yield= 𝑡ℎ𝑒𝑜𝑟𝑒𝑐𝑡𝑖𝑐𝑎𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 × 100% 1.4005𝑔

Percentage yield= 4.6237g ×100% = 30.24%

7) Draw the structure of the oxalatoaluminate anion (read the Introduction Section), taking oxalate as a bidentate ligand

Conclusion In this experiment, the mass of the final product, K3[Al (C2O4)3].3H2O is 1.4005g. The percentage yield of the product is 30.24% References 1) Al(OH)3 Aluminium hydroxide [Online] Available at: http://www.allreactions.com/index.php/group-3a/aluminium/aluminium-hydroxide 2) Oxidation-Reduction Reactions[Online] Available at: http://mmsphyschem.com/redox.pdf