A solution containing 0.095 mol. dm\(^{-3}\) of trioxonitrate (V) acid. Solution B contains 13.50g of X\(_2\)CO\(_3\).10H\(_2\)O per dm\(^3\)
(a) Put A in 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 B in mol. dm\(^3\)
(ii) molar mass of X\(_2\)CO\(_3\).10H\(_2\)O
(iii) percentage by mass of X in X\(_2\)CO\(_3\).10H\(_3\)O. The equation for the reaction is X\(_2\)CO\(_3\) + 2HNO\(_3{(aq)}\) \(\to\) 2XNO\(_{3(aq)}\) + CO\(_{2(g)}\) + 11H\(_2\)O\(_{(l)}\) [H = 1, C = 12, O = 16]
(c) Give the reason for the following:
(i) using just a small quantity of the indicator during acid-base titrations.
(ii) obtaining at least two sets of readings for titration experiment.
Explanation
Volume of pipette = 25cm\(^3\)
Indicator - Methyl orange
| Burette Readings | Rough | 1st | 2nd | 3rd |
| Final Burette Readings (cm\^3\)) | 25.20 | 29.90 | 27.00 | 25.00 |
| Initial Burette Reading (cm\(^3\)) | 0.00 | 5.05 | 2.20 | 0.10 |
| Volume of acid used (cm\(^3\)) | 25.20 | 24.85 | 24.80 | 24.90 |
(a) Average volume of A used = \(\frac{24.85 + 24.80 + 24.90}{3}\) = 24.85
The equation for the reaction \(\to\) X\(_2\)CO\(_3\).10H\(_2\)O\(_{(aq)}\) + 2HNO\(_{3(aq)}\) \(\to\) 2XNO\(_{3(aq)}\) + CO\(_{2(g)}\) + 11H\(_2\)O\(_{(l)}\)
(b)(i) Mole ratio of Acid to Base is 2:1
\(\frac{C_AV_A}{C_BV_B}\) = \(\frac{2}{1}\)
\(\frac{0.095 \times 24.85}{C_B \times 25}\) = \(\frac{2}{1}\)
C\(_B\) = \(\frac{0.095 \times 24.85}{2 \times 25} = 0.0472\) mol. dm\(^{-3}\)
(ii) Molar mass of X\(_2\)CO\(_3\).10H\(_2\)O = \(\frac{\text{mass concentration}}{\text{molar concentration}}\) = \(\frac{13.50}{0.0472}\) = 286 mol\(^{-1}\)
(iii) Relative molar mass of X\(_2\)CO\(_3\).H\(_2\)O = 286
2X + 12 + (16 x 3) + 10((2 x 1) + 16) = 286
2X + 240 = 286
2X = 46
Percentage by mass of X is X\(_2\)CO\(_3\).10H\(_2\)O = \(\frac{46}{280}\) x 100% = 16.08%
(c)(i) In order to obtain sharp and clear colour change at end point
(ii) For consistency or accuracy of titre values