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 require...
Question 1 Report
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 = \(22.4\text{ dm}^3\)]
Answer Details
(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.
Equation: Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)
(b)(i) Concentration of A
A contains 1.04 g HCl per 500 cm3 = 2.08 g dm-3. Molar mass of HCl = 1 + 35.5 = 36.5 g mol-1.
CA = 2.08 / 36.5 = 0.057 mol dm-3.
(b)(ii) Concentration of B
n(HCl) = 0.057 × 24.90 / 1000 = 1.42 × 10-3 mol. From the equation, n(Na2CO3) = half of n(HCl) = 7.10 × 10-4 mol in the 25.0 cm3 portion of B.
CB = 7.10 × 10-4 × 1000 / 25.0 = 0.0284 mol dm-3.
(b)(iii) Solubility of Na2CO3
B was made by diluting 50.0 cm3 of the saturated solution to 1000 cm3, a 20-fold dilution.
Solubility = 0.0284 × 20 = 0.568 mol dm-3.
(b)(iv) Volume of CO2 from 1 dm3 of B at s.t.p.
n(Na2CO3) in 1 dm3 of B = 0.0284 mol, and each mole gives 1 mole of CO2.
V(CO2) = 0.0284 × 22.4 = 0.64 dm3.
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