Burette readings (initial and final reading) must be given to two decimal places. Volume of pipette used must also be recorded but no account of experimental procedure is required. All calculations must be done in your answer book.
A is 0.0950 mol dm\(^{-3}\) HCI. B is a solution 13.50g dm\(^{-3}\) of X\(_2\)CO\(_3\).10H\(_2\)O.
(a) Put A into the burette and titrate it against 20.0 cm\(^3\) or 25.0 cm\(^3\) portions öf B using methyl orange as an indicator. Tabulate your readings and calculate the average volume of A used.
(b) From your results and the information provided above, calculate the;
(i) concentration of B in mol dm\(^{-3}\);
(ii) molar mass of X\(_2\)CO\(_3\).10H\(_2\)O in g mol\(^{-1}\);
(iii) percentage by mass X in X\(_2\)C)\(_3\).10H\(_2\)O. [H = 1, C = 12, O = 16]. The equation for the reaction involved in the titration is 2HCl\(_{(aq)}\) + X\(_2\)CO\(_3\).10H\(_2\)O\(_{(aq)}\) \(\to\) 2XCl\(_{(aq)}\) + 11H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\)
Explanation
Volume of pipette = 25.00 cm\(^{3}\)
indicator used = methyl orange
colour change at end point yellow to orange/purple
| Burette Reading | Rough | (1) First | (2nd) Second | (3rd) Third |
| Final (cm\(^3\)) | 25.20 | 24.90 | 24.80 | 24.82 |
| Initial (cm\(^3\)) | 0.00 | 0.00 | 0.00 | 0.00 |
| Volume of acid used | 25.20 | 24.00 | 24.80 | 24.85 |
Average volume of A used = \(\frac{24.90 + 24.80 + 24.85}{3}\) = 24.85 cm\(^3\)
The equation for the reaction
X\(_2\)CO\(_3\).10H\(_2\)O\(_{(aq)}\) + 2HNO\(_{3(aq)}\) \(\to\) 2 x 2NO\(_{3(aq)}\) + CO\(_{2(g)}\) + 11HNO\(_{(l)}\)
(b)(i) Mole ratio of Acid to Base is 2:1
concentration of B in mol. dm\(^{-3}\)
\(\frac{C_AV_A}{C_BV_B} = \frac{2}{1}\)
C\(_B\) = \(\frac{C_AV_A}{2V_B}\)
C\(_B\) = \(\frac{0.0950 \times 24.85}{2 \times 25}\)
= \(\frac{2.36075}{50}\)
= 0.0472 mol. dm\(^{-3}\)
(ii) Molar mass B in g mol\(^{-1}\)
Molar mass X\(_2\)CO\(3\).10H\(_2\)O = \(\frac{\text{mass concentration gdm}^{-3}}{\text{molar concentration of B in moldm}^{-3}}\)
= \(\frac{13.5 gdm^{-3}}{0.0472 moldm^{-3}}\)
= 286 g mol\(^{-1}\)
(iii) Percentage by mass of X
X\(_2\)CO\(_3\).10H\(_2\)O = 286g
i.e. relative molar mass of X\(_2\)CO\(_3\)
= 2x + 12 + (16 x 3) + 10(2 x 1) + 16 = 286
= 2x + 12 + 48 + 180 = 286
= 2x + 240 = 286
2x = 286 - 240 = 46
x = \(\frac{46}{2}\)
= 23 gmol\(^{-1}\)
Percentage by mass of X = \(\frac{2 \times 23}{286} \times \frac{100}{1}\)
= 16.08%