CHEMISTRY 3ThiChung Le

ThiChung Le Unit 3 Assignment: Solutions, solubility, and acid-base chemistry 1. For each of the following mixtures, explain whether it is a solution (homogeneous mixture) or a heterogeneous mixture. For the solutions only, identify the solute and the solvent. [5 marks] a. Ketchup: homogeneous, the solvent is water, and the solute is a combination of various ingredients in ketchup including tomatoes, vinegar, sugar, salt, and spices. b. Ice water: heterogenous, non-uniform composition. c. Orange juice: homogeneous,the solvent is water ,and the solute is orange ( which is include dissolved sugars, flavors, and other compounds present in the juice). d. Rainwater: homogeneous, the solvent is water, and the solute is sodium dissolved which contains in rainwater include gases, minerals, and particulate matter from the atmosphere. e. 14-karat gold in jewelry: homogeneous mixture. 2. Suppose to you eat a piece of candy and let it dissolve it in your mouth. Identify the solvent and solute for the resulting solution? [4 marks] When you eat a piece of candy and it dissolves in your mouth, the solvent is saliva and the solute is the candy. The solvent: - Watery solution is Saliva which contains enzymes, electrolytes, and other substances. - The main purpose of this components is help to break down the candy become smaller pieces that can be dissolved and absorbed by the body. - In the process of lubricating the food and making it easier to swallow, Saliva is also play an important role into it. The solute: - There are various ingredients, including sugars, starches, fats and other flavorings that candy contains. - These ingredients begin to dissolve and break down, when the candy comes into meet with saliva. - The dissolved candy components and the components of saliva mixture is the resulting of the solution.

3. Write reactions to show the dissociation into ions of the following substances when they dissolve in water: [4 marks] a. Sodium phosphate: Na 3 PO 4 Na 3 PO 4 is a strong electrolyte and It dissociates completely into ions ( 3Na + and 1PO 4 3- ) when it dissolves in water. Na 3 PO 4 (s) → 3Na + (aq)+ PO 4 3- (aq) b. Ammonium hydroxide: NH 4 OH NH 4 OH is d issolves in water and i t dissociates completely into ions ( 1NH 4 + and 1OH - ) NH 4 OH(aq) → NH 4 + (aq) + OH - (aq) 4. Explain why water is known as the “universal solvent.” [4 marks] Because of the polarity which make water become excellent solvent . In water, Oxygen atoms tend to attract more electrons than hydrogen atoms. Because of that, the molecule has a partially positive end and a partially negative end. At the result, polar lysing molecules can be separated by water molecules. The solute molecules is surrounded by the solvent molecules when the solute dissolves in water. Water molecules are generally small, so a lot of them can surround the molecules in solution Polar covalent bonds have charged ends, but Ionic material split into charge ions. This help water to easily dissolve such substances. At the result, ionic substances and polar covalent bonds are easily dissolved in water. 5. Use a solubility table to predict which of the following combinations of compounds would produce a precipitate. State what the precipitate will be if there will be one. (4 marks) a. strontium hydroxide and ammonium phosphate: The products in this reaction are the result of the exchange of ion partners. That is, the strontium ion combine with the phosphate ion, and the ammonium ion combines with the hydroxide ion. The latter combination results in NH 4 OH, which is a soluble compound. Therefore, NH 4 + and OH − remain dissolved in the solution. However, Sr 2+ and PO4 3− combine to produce Sr 3 (PO4) 2 , which is an insoluble compound. Therefore, The combination of the compounds results in the precipitation of Sr (PO ) . ₃ ₄ ₂

Since nitrous acid is a weaker acid, it has fewer ions in the solution, resulting in lower conductivity. On the other hand, nitric acid is a strong acid, resulting in more ions in the solution and thus higher conductivity. 8. Identify the following household substances as either very acidic, slightly acidic, very basic, slightly basic, or neutral. [5 marks] 1. drain cleaner, pH = 13 Very basic 2. milk, pH = 6.8 neutral 3. baking soda, pH = 10 Slightly basic 4. coffee, pH = 5 Slightly acidic 5. muriatic acid, pH = 1 Very acidic 9. Explain the Arrhenius theory of acids in your own words. Write a reaction showing why hydrogen carbonate is an acid. [4 marks] According to Arrhenius theory, an acids is a substance that dissociates to give H+ ions in an aqueous medium. Example: HNO3- is an Arrhenius acid. HCO 3 - (aq) ⇌ H + (aq)+ CO 3 2- (aq) In this reaction, hydrogen carbonate (HCO3-) dissociates in water to produce a hydrogen ion (H+) and a carbonate ion (CO3^2-). The hydrogen ion (H+) released is responsible for the acidic behavior of hydrogen carbonate. 10. Determine the concentration as a percentage by volume if a 500 mL bottle of vinegar contains 40.0 mL of acetic acid. [2 marks] We have: V solute = 40 ml ( acetic acid) V solution = 500 ml ( vinegar) C = ?

C = (V solute / V solution ) x 100% C = ( 40 ml / 500 ml ) x 100% C = 8 % Therefore, the concentration as a percentage by volume if a 500 mL bottle of vinegar contains 40.0 mL of acetic acid is 8%. 11.Determine the concentration as a percentage weight by volume if a 250 mL aqueous sodium chloride solution contains 22.7 g of NaCl. [2 marks] We have: m solute = 22.7 g ( NaCl ) V solution = 250 ml (aqueous sodium chloride ) C = ? C = ( m solute / V solution ) x 100% C = ( 22.7 g / 250 ml ) x 100% C = 9.08 % Therefore, he concentration as a percentage weight by volume if a 250 mL aqueous sodium chloride solution contains 22.7 g of NaCl is 9.08 %. 12. Calculate the molar concentration of a 6.70 L solution that contains 38.5 g of dissolved sodium hydrogen carbonate. [6 marks] We have: mNaHCO3 = 38.5 g MNaHCO3 = MNa+ MH+MC+3xMO MNaHCO3 = 23.00 g/mol+ 1 g/mol + 12.01g/mol + 3 x 15.999 g/mol MNaHCO3 = 84.0 g/mol So nNaHCO3 = mNaHCO3 / MNaHCO3 nNaHCO3 = 38.5( g) / 84.0 (g/mol) nNaHCO3 = 0.458 mol We have: Vsolution = 6.7 L So C = n / V = 0.458 mol / 6.7 L C = 0.068 Therefore, the molar concentration is 0.068 13.You and your friend conduct an experiment in which you need to make 150 mL of a 0.250 M NaOH solution from an existing stock solution which has a concentration 6.25 M. What volume of concentration sodium hydroxide is required? [7 marks] We have: V2 = 150 ml = 0.15 L

1- 2 ml dilute sulphuric acid (H 2 SO 4 ) solution Test tubes Droppers Stirring rod Centrifuge or filtration setup Experimental Setup: Preliminary test: To a small quantity of the given sample salt, 1−2 mL of dilute H2SO4 is added and note the changes. Observations Inference Gas evolved Possible Anion A colourless, odouless gas is evolved with brisk effervescence, which turns lime water milky. CO2 Carbonate (CO32-) Colourless vapours with smell of vinegar. Vapours turn blue litmus red. CH3COOH vapours Acetate, (CH3COO-) Confirmatory test Procedure Observations Inference Take 0.1 g of salt in a test tube, add dilute sulphuric acid CO2 is evolved which turns lime water milky. Presence of carbonate. On passing the gas for some more time, milkiness disappears. Carbonate is confirmed. Take 0.1 g of salt in a china dish. Add 1 mL of ethanol and 0.2 mL conc. H2SO4 and heat Fruity odour Presence of acetate Take 0.1 g of salt in a test tube, add 1-2 mL distilled water. Shake well filter if necessary. Add 1-2 mL Deep red colour appears which disappears on boiling and a brown-red precipitate is formed. Acetate is confirmed.

neutral ferric chloride solution to the filtrate. TEST FOR CARBONATE ION: Adding dil.H 2 SO 4 to the carbonate salt: Na 2 CO 3 + H 2 SO 4 Na ⟶ 2 SO 4 + H 2 O + CO 2 Formula of lime water Ca(OH) 2 On passing CO 2 to limewater, Ca(OH) 2 + CO 2 CaCO ⟶ 3 + H 2 O The lime water turns milky due to the formation of Calcium carbonate. On passing the gas for some more time, CaCO 3 + H 2 O + CO 2 Ca(HCO ⟶ 3 ) 2 The milkiness disappears due to the formation of Calcium hydrogen carbonate which is soluble in water. TEST FOR ACETATE ION : Adding H 2 SO 4 and Ethanol to the acetate salt: 2CH 3 COONa + H 2 SO 4 Na ⟶ 2 SO 4 + 2CH 3 COOH Adding H2SO4 results in a vinegar smell due to the presence of acetate CH 3 COOH+C 2 H 5 OH CH ⟶ 3 COOC 2 H 5 Adding C2H5OH results in a fruity order due to the presence of ethyl acetate (ester). Adding neutral FeCl3 solution to acetate ion giving deep red colour. 6CH 3 COO - + 3Fe 3+ + 2H 2 O [Fe(OH) ⟶ 2 (CH 3 COO) 6 ] + + 2H + Red colur is due to the formation of this complex ion. Then on boiling this forms, [Fe(OH) 2 (CH 3 COO) 6 ] + + 4H 2 O 3[Fe(OH) ⟶ 2 (CH 3 COO)] + 3CH 3 COOH+ H +

16. Calculate the pH of a hydrochloric acid solution with a concentration of 2.35×10-6 mol/L [2 marks] We have : pH = -log[H+] pH = -log(2.35 x 10 -6 ) pH = -6.37 pH = -6.4 17. For this question you will be writing a properly formatted lab report with headings for two sections. Your task is to use the observations and data provided to complete the results section and perform calculations. (Note: You do not need to complete any other required sections) [10 marks] We have: V HC2H3O2trial1 = 24.27 - 0.00 = 24.27ml V HC2H3O2trial2 = 48.55mL - 24.27mL = 24.28ml V HC2H3O2trial3 = 72.81mL - 48.55mL = 24.26 ml V HC2H3O2 = (V HC2H3O2trial1 +V HC2H3O2trial2 +V HC2H3O2trial3 ) / 3 V HC2H3O2 = ( 24.27ml + 24.28 ml + 24.26 ml) / 3 V HC2H3O2 = 24.27 ml

NaOH (aq) + HC 2 H 3 O 2 (aq) → NaC 2 H 3 O 2 (aq) + H 2 0 ( l) We have: C NaOH = 2.5 x 10 -4 mol/L V= 24.27 ml = 0.024 L So n NaOH = C x V n NaOH = 0.0242 L x ( 2.5 x 10 -4 ) mol/L n NaOH = 0.0605 x 10 -4 mol/L We have the ratio of NaOH (aq) and HC 2 H 3 O 2(aq) is 1:1 n HC2H3O2 = n NaOH x( 1 /1 ) = 0.0605 x 10 -4 mol/L We use 10 ml of HC 2 H 3 O2 in the experiment, so V= 10 (ml) = 0.01 (L) So C HC2H3O2 = n / V C HC2H3O2 = 0.0605 x 10 -4 (mol/l) / 0.01 (L) C HC2H3O2 = 6.05 x 10 -4