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 containing 14.0gdm\(^{-3}\) of potassium hydrogen tetraoxosulphate (VI). B is a solution of sodium hydroxide.
(a) Put A into the burette and titrate with 20.0cm\(^{-3}\) or 25.0 cm\(^{-3}\) portion of B using methyl orange, as 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 A in mol dm\(^{3}\)
(ii) concentration of B in g dm\(^3\)
(iii) volume of A (in dm\(^{-3}\)) that would produce one mole of Sodium tetraoxosulphate (VI) in solution. The equation for the reaction is:
2KHSO\(_{4(aq)}\) + 2NaOH\(_{(aq}\) \(\to\) K\(_2\)SO\(_{4(aq)}\) + NaSO\(_{4(aq)}\) + 2H\(_2\)O\(_{(l)}\) [H = 1; O = 16, Na = 23, KHSO\(_4\) = 136gmol\(^{-2}\)]
| Burette readings | 1st Rough | 2nd Rough | 3rd Rough |
| Final burette reading in cm\(^3\) | 24.80 | 24.00 | 26.00 |
| Initial burette reading cm\(^3\) | 0.00 | 0.00 | 2.00 |
| Difference in cm\(^{3}\) | 24.80 | 24.00 | 24.00 |
(a) Average volume of A used = \(\frac{24.00 + 24.00}{2}\)
= 24.00 cm
Volume of pipette (i.e) B = 25.00cm
(b)(i) Concentration of A = \(\frac{140}{136}\) 0.103 mol dm\(^3\)
(ii) The equation for the reaction
2KHSO\(_4\) + 2NaOH \(\to\) K\(_2\)SO\(_4\) Na\(_2\)SO\(_4\) + 2H\(_2\)O; From equation mole ratio is 1:1 or 2:2
\(\frac{C_AV_A}{C_BV_B}\) = 1
Where C\(_A\) is the concentration of A,
A\(_A\) is the volume of A,
C\(_B\) is the concentration of B
V\(_B\) is the volume of B
C\(_B\) = \(\frac{0.103 \times 23}{C_B \times 25}\)
= 40 x C\(_B\) = 40 x C\(_B\)
= 40 x 0.0948
= 3.79 g/dm\(^3\)
(iii) From the equation, 2 mole KHSO\(_4\) will produce or yield 1 mole Na\(_2\)SO\(_4\)
1 dm\(^3\) of A contains 0.103 mole - volume of A that contains 2 moles KHSO\(_4\) = \(\frac{2}{0.103}\) = 19.4 dm\(^3\)
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