All your burette readings (initial and final), as well as the size of your pipette, must be recorded but on no account of experimental procedure is required. All calculations must be done in your answer book.
A is 0.50 mol dm\(^{-3}\) hydrochloric acid. B is 0.025 mol dm\(^{-3}\) of a trioxocarbonate (IV) salt.
(a) Put A into the burette and titrate with 20.0cm\(^{-3}\) or 25.0 cm\(^{-3}\) portions of B using methyl orange or screened methyl Orange indicator. Record the volume of your pipette. Tabulate your burette readings and calculate the average of A used.
(b) From your results, calculate the mole ratio of acid to trioxocarbonate (IV) in the reaction, expressing your answer as a whole number ratio of one.
(c) Given that B contains 7.2g dm\(^{-3}\) of the hydrated trioxocarbonate (IV) salt, calculate the:
(i) concentration of anhydrous salt in B in g dm\(^{-3}\) [Molar mass of anhydrous salt in B = 106g]
(ii) percentage of water of crystallization in the hydrated salt.
Explanation
| Burette readings | Rough | 1st | 3rd | 3rd |
| Final burette reading (cm\(^3\)) | 23.50 | 33.00 |
23.00 |
28.00 |
| Initial burette reading (cm\(^3\)) | 0.00 | 10.00 | 0.00 | 5.00 |
| Initial burette reading (cm\(^3\)) | 23.50 | 23.00 | 23.00 | 23.00 |
(a) Average volume of the acid(titre) = \(\frac{23.00 + 23.00 + 23.00}{3}\) = 23.00 cm\(^3\)
(b) A is 0.050 mol dm\(^{-3}\) of HCl; B is 0.25 mol dm\(^{-3}\) of XCO\(_3\)
Let the mole ratio of the acid to the triocarbonate (IV) be M
From \(\frac{\text{concentration of A(C\(_A\)) x volume of (V\(_A\))}}{\text{concentration of B(C\(_B\)) x volume of (B)(V\(_B\))}}\)
\(\frac{0.050 \times 23.00}{0.025 \times 25.00}\) = m
m = 1.84
Since m = \(\frac{A}{B}\)
A : B = 1.84 : 1
A : B approximately 2 : 1 expressed in a whole number ratio
(c) Molar mass of XCO\(_3\) = 160g (given); concentration = 7.2g/dm\(^3\) (hydrated trioxocarbonate)
(i) concentration of anhydrous salt = mole concentration x molar mass
0.025 x 106 = 2.65 m/dm\(^3\)
(ii) percentage of water of crystallisation = \(\frac{7.2 - 2.65}{7.2} \times \frac{100}{1}\)%
= \(\frac{4.55}{7.2}\) x 100%
= 63.19%