All your burette readings (initials and final) as well as the size of your pipette must be recorded but no account experimental procedure is required. All calculations must be done in your answer booklet.
A is 0.100 mol dm\(^{-3}\) HNO\(_3\). B is a solution containing 2.50 g of a mixture of Na\(_2\)CO\(_3\) and Na\(_2\)SO\(_4\) in 250 cm\(^3\) of solution.
(a) Put into the burette and titrate it against 20.0 cm\(^3\) or 25.0cm\(^3\) portions of B using methyl orange as indicator. Repeat the exercise to obtain consistent titre values. Tabulate your readings and calculate the average volume of acid A used. The equation of reaction is Na\(_2\)CO\(_{3(aq)}\) + 2HNO\(_{3(aq)}\) \(\to\) NaNO\(_{3(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\)
(b) From your results and the information provided, calculate the:
(i) concentration of B in moldm\(^{-3}\);
(ii) concentration of Na\(_2\)CO\(_3\) in B in gdm\(^{-3}\);
(iii) percentage of Na\(_2\)CO\(_3\), in the mixture. [Na\(_2\)CO\(_3\) = 106]
Explanation
(a)
| Burette reading | Rough | 1st titration | 2nd titration |
|
Final Burret Reading (cm\(^3\))
|
23.50 | 45.90 | 33.60 |
|
Initial Burret Reading (cm\(^3\)) |
01.10 | 23.50 | 11.30 |
|
Volume of acid used (cm\(^3\)) |
22.40 | 22.40 | 22.30 |
Average titre = \(\frac{22.40 + 22.40 + 22.30}{3}\) = 22.37 cm\(^3\)
(b)(i) Conc. of B in mol dm\(^{-3}\)
\(\frac{C_AV_A}{C_BV_B}\) = \(\frac{2}{I}\)
C\(_B\) = \(\frac{C_AV_A}{2VB}\)
= \(\frac{0.100 \times 22.37}{2 \times 25}\)
= 0.0447 mol dm\(^{-3}\)
(ii) Conc. of Na\(_2\)CO\(_3\) in Bg dm\(^{-3}\)
Conc. of Na\(_2\)CO\(_3\) = 0.0447 x Molar Mass of B
= 0.0447 x 106
= 4.74 g dm\(^{-3}\)
(iii) 250 cm\(^3\) of B contained 2.50 g of Na\(_2\)CO\(_3\)
1000 cm\(^3\) of B = \(\frac{2.50 \times 1000}{250}\)
= 10.0g
mass conc of B = 10 g dm\(^{-3}\)
% Na\(_2\)CO\(_3\) = \(\frac{4.74}{10} \times 100\)
= 47.4%