A is a solution containing 0.050 mol. dm of tetraoxosulphate (VI) acid. B is a solution of anhydrous trioxocarbonate (IV).
(a) Put A into the burette and titrate with 20cm\(^3\) or 25cm\(^3\) 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
(b) From your results and the information provided, calculate the:
(i) Concentration of solution B in mol. dm\(^{-3}\)
(ii) mass of sodium tetraoxosulphate (VI) that would be formed in solution of 1dm\(^3\) of solution B were neutralized by solution A
(iii) volume of carbon (IV) oxide at s.t.p. that would be liberated in (b)(ii) above. The equation for the reaction is: N\(_2\)CO\(_{3(aq)}\) + H\(_2\)SO\(_{4(aq)}\) \(\to\) Na\(_2\)SO\(_{4(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\) [O = 16; Na = 23, S = 32; molar volume of gases of s.t.p. = 22.4dm\(^3\)
Volume of pipette = 25 cm\(^3\)
Indicator - methyl orange
| Burette readings | Rough | 1st | 2nd | 3rd |
| Final burette readings (cm\(^3\)) | 23.10 | 45.00 | 23.60 | 22.40 |
| Initial burette readings (cm\(^3\)) | 0.50 | 22.60 | 1.20 | 0.00 |
| burette reading (cm\(^3\)) | 22.60 | 22.40 | 22.40 | 22.40 |
(i) Average volume of A used = \(\frac{22.40 + 22.40 + 22.40}{3}\) = 22.40cm\(^3\)
(b)(i) Equation for the reaction, mole ratio of acid to base (salt) is 1 : 1
\(\frac{\text{Molar concentration of A(C\(_A\)) x volume of A(V\(_A\))}}{\text{Molar concentration of B(C\(_B\)) x volume of B(V\(_B\))}}\) = mole reaction of \(\frac{A}{B}\) = \(\frac{0.05 \times 22.40}{C_B \times 25} = \frac{1}{1}\)
C\(_B\) = \(\frac{0.05 \times 22.40 \times 1}{25}\) = 0.0448 mol. dm\(^3\)
(ii) Molar mass of Na\(_2\)SO\(_4\) = (23 x 2) + 32 + (16 x 4) = 142g
Amount of Na\(_2\)CO\(_3\) in 1 dm\(^3\) of B = 0.0448 mole
from equation
1 mole of Na\(_2\)CO\(_3\) = 0.0488 x 142 = 6.36g
(iii) From the equation, 1 mole of Na\(_2\)CO\(_3\) produces 1 mole of CO\(_2\) at s.t.p.
i.e. mole of Na\(_2\)CO\(_3\) produces 22.4 dm\(^3\) of CO\(_2\) at s.t.p
Mole of Na\(_2CO_3\) produces \(\frac{22.4}{1}\) x 0.0448 = 1.00 dm\(^3\)
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