All your burette readings (initial and final), as well as the 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 of H\(_2\)SO\(_4\) containing 4.9 gdm-3, B is a solution containing X g dm\(^{-3}\) of Na\(_2\)CO\(_3\).
(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 an indicator. Record the volume of your pipette. Tabulate your burette readings and calculate the average volume of A used. The equation for the reaction involved in the titration is; H\(_2\)SO\(_{4(aq)}\) + Na\(_2\)CO\(_{3(aq)}\) \(\to\) Na\(_{2}\)SO\(_{4(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\)
(b) From your results and information provided above, calculate the:
(i) Concentration of A In mol dm\(^{-3}\)
(ii) concentration of B in mol dm\(^{-3}\)
(iii) mass of salt formed when 500 cm\(^3\) of B is Completely neutralized by A.
(v) volume of carbon (IV) oxide liberated in (b) (ii) above at s.t.p. [O = 16, Na = 23, S = 32, 1 mole or a gas occupies 22.4 dm\(^3\) at s.t.p.]
Explanation
| Titration/Burette Reading | Rough(cm\(^3\)) |
1st Titre (cm\(^3\))
|
2nd Titre (cm\(^3\) | 3rd Titre (cm\(^3\)) |
|
Final Reading |
22.80 | 22.70 | 32.70 | 22.70 |
| Initial Reading | 0.00 | 0.00 | 10.00 | 0.00 |
| Volume of acid used | 22.00 | 22.70 | 22.70 | 22.70 |
Volume of pipette used = 25.00 cm\(^3\)
Average volume of acid = \(\frac{22.70 + 22.70 +22.70}{3}\)
= \(\frac{68.10}{3}\)
= 22.70 cm\(^3\)
(b)(i) Concentration of A in mol/dm\(^3\) = \(\frac{\text{conc. of A in g/dm}^3}{\text{Molar mass of A}}\)
Molar mass of H\(_2\)SO\(_{4(aq)}\) = 1 x 2 + 32 x 16 x 4 = 98 g/mol
Conc. of A in mol/dm\(^3\) = \(\frac{4.9}{98}\) = 0.05 mol/dm\(^3\)
(ii) H\(_2\)SO\(_4\) + Na\(_2\)CO\(_{3(aq)}\) \(\to\) Na\(_2\)SO\(_{4(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\)
\(\frac{C_AV_A}{C_BV_B} = \frac{\text{Mole ratio of acid}}{\text{Molar ration of base}}\)
\(\frac{0.05 \times 22.7}{C_B \times 25} = \frac{1}{1}\)
C\(_B\) = \(\frac{0.05 \times 22.7 \times 1}{25 \times 1}\)
= 0.0454
C\(_B\) = 0.05 3 S.F
C\(_B\) = 0.05 mol/dm\(^3\)
(iii) Moles of B = \(\frac{ \text{conc. of B in mol/dm}^3 \times Volume of B}{1000}\)
\(\frac{0.05 \times 500}{1000}\) = 0.025 moles
Molar mass of Na\(_2\)SO\(_4\) = 2 x 23 + 32 + 16 x 4 = 142 g/mol
1 mole of B formed 142g of Na\(_2\)SO\(_4\)
0.025 moles of B will form \(\frac{0.025 \times 142}{1}\) = 3.55 g
(iv) 1 mole of B produces 22.4 dm\(^3\) of CO\(_2\) at s.t.p.
0.025 mole of B will produce 0.025 x 22.4 = 0.56 dm\(^3\) of CO\(_2\) at s.t.p. C is CaCl\(_2\), D is CH\(_3\)COOH or Benzoic acid