TEST OF PRACTICAL KNOWLEDGE QUESTION
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.]
(a) Burette readings and average titre
Volume of pipette used = 25.00 cm3 of B.
| Burette reading (cm3) | Rough | 1st titre | 2nd titre | 3rd titre |
|---|
| 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 |
Average volume of A used (concordant titres):
\[ V_A = \frac{22.70 + 22.70 + 22.70}{3} = \frac{68.10}{3} = 22.70\ \text{cm}^3. \]
(b)(i) Concentration of A in mol dm-3
Molar mass of H2SO4 = \(2(1) + 32 + 4(16) = 98\ \text{g mol}^{-1}\).
\[ C_A = \frac{4.9}{98} = 0.05\ \text{mol dm}^{-3}. \]
(b)(ii) Concentration of B in mol dm-3
From the equation the acid : base mole ratio is 1 : 1, so \(\dfrac{C_A V_A}{C_B V_B} = \dfrac{1}{1}\):
\[ C_B = \frac{C_A V_A}{V_B} = \frac{0.05 \times 22.70}{25.0} = 0.0454 \approx 0.05\ \text{mol dm}^{-3}. \]
(b)(iii) Mass of salt (Na2SO4) formed by 500 cm3 of B
Moles of Na2CO3 in 500 cm3 of B \(= \dfrac{0.05 \times 500}{1000} = 0.025\ \text{mol}\).
From the equation, moles of Na2SO4 formed = 0.025 mol. Molar mass of Na2SO4 \(= 2(23) + 32 + 4(16) = 142\ \text{g mol}^{-1}\).
\[ \text{Mass of Na}_2\text{SO}_4 = 0.025 \times 142 = 3.55\ \text{g}. \]
(b)(v) Volume of CO2 liberated at s.t.p.
Moles of CO2 = moles of Na2CO3 reacted = 0.025 mol.
\[ V_{CO_2} = 0.025 \times 22.4 = 0.56\ \text{dm}^3\ \text{at s.t.p.} \]
(a) Burette readings and average titre
Volume of pipette used = 25.00 cm3 of B.
| Burette reading (cm3) | Rough | 1st titre | 2nd titre | 3rd titre |
|---|
| 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 |
Average volume of A used (concordant titres):
\[ V_A = \frac{22.70 + 22.70 + 22.70}{3} = \frac{68.10}{3} = 22.70\ \text{cm}^3. \]
(b)(i) Concentration of A in mol dm-3
Molar mass of H2SO4 = \(2(1) + 32 + 4(16) = 98\ \text{g mol}^{-1}\).
\[ C_A = \frac{4.9}{98} = 0.05\ \text{mol dm}^{-3}. \]
(b)(ii) Concentration of B in mol dm-3
From the equation the acid : base mole ratio is 1 : 1, so \(\dfrac{C_A V_A}{C_B V_B} = \dfrac{1}{1}\):
\[ C_B = \frac{C_A V_A}{V_B} = \frac{0.05 \times 22.70}{25.0} = 0.0454 \approx 0.05\ \text{mol dm}^{-3}. \]
(b)(iii) Mass of salt (Na2SO4) formed by 500 cm3 of B
Moles of Na2CO3 in 500 cm3 of B \(= \dfrac{0.05 \times 500}{1000} = 0.025\ \text{mol}\).
From the equation, moles of Na2SO4 formed = 0.025 mol. Molar mass of Na2SO4 \(= 2(23) + 32 + 4(16) = 142\ \text{g mol}^{-1}\).
\[ \text{Mass of Na}_2\text{SO}_4 = 0.025 \times 142 = 3.55\ \text{g}. \]
(b)(v) Volume of CO2 liberated at s.t.p.
Moles of CO2 = moles of Na2CO3 reacted = 0.025 mol.
\[ V_{CO_2} = 0.025 \times 22.4 = 0.56\ \text{dm}^3\ \text{at s.t.p.} \]
