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 g HCl per 500 cm\(^3\) of solution. B was prepared by diluting 50.0 cm\(^3\) of a saturated solution of Na\(2\)CO\(_3\) at room temperature to 1000 cm\(^3\)
(a) Put A into the burette and titrate it against 20.0 cm\(^3\) or 25.0 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 moldm\(^{-3}\)
(ii) concentration of B in mol dm\(^{-3}\)
(iii) solubility of Na\(_2\)CO\(_3\) in mol dm\(^{-3}\)
(iv) volume of CO\(_2\) that would be liberated from 1 dm\(^3\) of B if the titration were carried out at s.t.p.
The equation for the reaction is Na\(_2\)CO\(_{3(aq)}\) + 2HCl\(_{(aq)}\) \(\to\) 2NaCl\(_{(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\)
[H = 1; C = 12; O = 16; Na = 23; Cl = 35.5; Molar volume of gas at s.t.p = 22.4 dm\(^3\)]
Explanation
(a)
| Burette Reading |
Rough Titration (cm\(^3\)) |
1st Titration (cm\(^3\)) |
2nd Titration (cm\(^3\)) | 3rd Titration (cm\(^3\)) |
| Final | 25,00 | 24.90 | 34.90 | 24.90 |
| Initial | 0.00 | 0.00 | 10.00 | 0.00 |
| Volume of acid used | 25.00 | 24.90 | 24.90 | 24.90 |
Average volume of acid used
= 24.90 + \(\frac{24.90}{3}\) + 24.90 = \(\frac{74.7}{3}\)
= 24.90 cm\(^3\)
(b) Concentration of A in g/dm\(^3\) = \(\frac{1.04 \times 1000}{500}\)
= 2.08 g/dm \(^3\)
Molar mass of HCl = 1 + 35.5 = 36.5 g/mol
Concentration of A in mol/dm\(^3\) = \(\frac{\text{concentration in g/dm}^3}{\text{molar mass}}\)
= \(\frac{2.08}{36.5}\)
= 0.057 mol/dm\(^3\)
(ii) \(\frac{C_AV_A}{C_BV_B} = \frac{N_A}{N_B}\)
\(\frac{0.057 \times 24.90}{C_B \times 25} = \frac{2}{1}\)
C\(_B\) = \(\frac{0.057 \times 24.90}{25 \times 2}\) = \(\frac{1.4193}{50}\)
= 0.02839
= 0.0284 mol/dm\(^3\)
(iii) To calculate the solubility of Na\(_2\)CO\(_3\) in mol/dm\(^3\). The dilution factor of B = 50 to 1000
= 1 : 20
The solubility = 0.0284 x 20
= 0.568 mol/dm\(^3\)
(iv) To calculate the volume of CO\(_{2(g)}\) liberated from 1 dm\(^3\) at s.t.p. from the equation of the reaction, one mole of Na\(_2\) CO\(_3\) liberate 22.4 x 0.0284
= 0.64 dm\(^3\) of CO\(_2\)