Credit will be given for strict adherence to the instructions, for observations precisely recorded, and for accurate inferences. All tests, Observations, and inferences must be clearly entered in your answer book, in ink, at the time they are made.
C is a mixture of two inorganic salts. Carry out the following exercises on C. Record your observations and identify any gas(es) evolved. State the conclusion you draw from the result of each test.
(a) Put all of C in a test tube and add about \(10\ \text{cm}^3\) of distilled water. Stir, filter, and keep the filtrate and the residue.
(b) Put the residue into a test tube and add about \(5\ \text{cm}^3\) of dilute HCl. Shake to dissolve
(i) To about \(2\ \text{cm}^3\) of the solution, add \(\mathrm{NaOH_{(aq)}}\) in drops and then in excess
(ii) To another \(2\ \text{cm}^3\) portion of the solution, add \(\mathrm{NH_{3(aq)}}\) in drops and then in excess.
(c) To about \(2\ \text{cm}^3\) portion of the filtrate, add few drops of dilute \(\mathrm{HNO_3}\), and then \(\mathrm{AgNO_{3(aq)}}\) a followed by aqueous \(\mathrm{NH_3}\) in excess.
Results for specimen C
Test
Observation
Inference
(a) Add about 10 cm3 of distilled water to all of C, stir and filter.
C dissolves partly. A white residue remains on the filter paper and a colourless filtrate is obtained.
C contains a water-soluble salt and a water-insoluble white salt.
(b) Add dilute HCl to the residue and shake.
Effervescence occurs and the residue dissolves, giving a colourless solution. The colourless gas turns limewater milky.
The gas is carbon dioxide, CO2. A carbonate ion, CO32−, is present in the insoluble salt.
(b)(i) To 2 cm3 of the solution from (b), add NaOH(aq) dropwise and then in excess.
A white gelatinous precipitate forms. The precipitate dissolves in excess NaOH(aq).
Zn2+, Al3+ or Pb2+ may be present.
(b)(ii) To a fresh 2 cm3 portion of the solution from (b), add NH3(aq) dropwise and then in excess.
A white gelatinous precipitate forms. The precipitate dissolves in excess aqueous ammonia.
Zn2+ is confirmed.
(c) To 2 cm3 of the filtrate, add dilute HNO3, then AgNO3(aq), followed by excess NH3(aq).
There is no visible reaction with dilute HNO3. A white precipitate forms on adding AgNO3(aq); it dissolves in excess NH3(aq).
Cl− is present and is confirmed.
Conclusion: C is a mixture of insoluble zinc carbonate, ZnCO3, and a soluble chloride salt.
(a) State an indicator suitable for the titration of;
(i) dilute HCl and NaOH\(_{3(aq)}\)
(ii) dilute CH\(_3\)COOH and KOH\(_{(aq)}\)
(iii) dilute HCl and NH\(_{3(aq)}\).
Give a reason for your answer in each case.
(b) Calculate the volume of water that would be added to 50 cm\(^3\) of 0.10 mol dm\(^{-3}\) of HCI to dilute it to 0.010 mol dm \(^{-3}\)
(c) Name one gas that could be used to demonstrate the fountain experiment.
(a) Suitable indicators for the titrations
(i) Dilute HCl (strong acid) and NaOH (strong base): either methyl orange or phenolphthalein is suitable. Reason: the salt formed (NaCl) is neutral and there is a large, sharp pH change at the end point that spans the ranges of both indicators.
(ii) Dilute CH3COOH (weak acid) and KOH (strong base): use phenolphthalein. Reason: the salt formed is of a weak acid and a strong base, so it hydrolyses to give an alkaline solution (pH > 7) at the end point, within the colour-change range of phenolphthalein.
(iii) Dilute HCl (strong acid) and NH3 (weak base): use methyl orange. Reason: the salt formed is of a strong acid and a weak base, so it hydrolyses to give an acidic solution (pH < 7) at the end point, within the range of methyl orange.
(i) Dilute HCl (strong acid) and NaOH (strong base): either methyl orange or phenolphthalein is suitable. Reason: the salt formed (NaCl) is neutral and there is a large, sharp pH change at the end point that spans the ranges of both indicators.
(ii) Dilute CH3COOH (weak acid) and KOH (strong base): use phenolphthalein. Reason: the salt formed is of a weak acid and a strong base, so it hydrolyses to give an alkaline solution (pH > 7) at the end point, within the colour-change range of phenolphthalein.
(iii) Dilute HCl (strong acid) and NH3 (weak base): use methyl orange. Reason: the salt formed is of a strong acid and a weak base, so it hydrolyses to give an acidic solution (pH < 7) at the end point, within the range of methyl orange.
All your burette readings (initial and final) as well as the size size of your pipette, must be recorded but no account of experimental procedure is required. All calculations must be done in your answer book.
A is a solution containing \(1.04\text{ g HCl}\) per \(500\text{ cm}^3\) of solution. B was prepared by diluting \(50.0\text{ cm}^3\) of a saturated solution of \(\mathrm{Na}_2\mathrm{CO}_3\) at room temperature to \(1000\text{ cm}^3\)
(a) Put A into the burette and titrate it against \(20.0\text{ cm}^3\) or \(25.0\text{ cm}^3\) portions of B using methyl orange as indicator. Repeat the titration to obtain consistent titres. Tabulate your results and calculate the average volume of acid used.
9b) From your results and information provided above, calculate the;
(i) concentration of A in \(\mathrm{mol\,dm}^{-3}\)
(ii) concentration of B in \(\mathrm{mol\,dm}^{-3}\)
(iii) solubility of \(\mathrm{Na}_2\mathrm{CO}_3\) in \(\mathrm{mol\,dm}^{-3}\)
(iv) volume of \(\mathrm{CO}_2\) that would be liberated from \(1\text{ dm}^3\) of B if the titration were carried out at s.t.p.
The equation for the reaction is \(\mathrm{Na}_2\mathrm{CO}_{3(aq)} + 2\mathrm{HCl}_{(aq)} \to 2\mathrm{NaCl}_{(aq)} + \mathrm{H}_2\mathrm{O}_{(l)} + \mathrm{CO}_{2(g)}\)
[H = 1; C = 12; O = 16; Na = 23; Cl = 35.5; Molar volume of gas at s.t.p = \(22.4\text{ dm}^3\)]
(a) Titration results
Rough
1
2
3
Final reading (cm3)
25.00
24.90
34.90
24.90
Initial reading (cm3)
0.00
0.00
10.00
0.00
Volume of A used (cm3)
25.00
24.90
24.90
24.90
Average volume of acid A used = (24.90 + 24.90 + 24.90) / 3 = 24.90 cm3.
