A is a solution containing 0.50 mole of barium chloride per dm\(^3\). Solution B contains 1.0 mole of tricxocarbonate (IV) salt per dm\(^3\)
(a) State what would be observed and give the confirmatory test for any gases evolved if the following tests were performed
(i) mixing 2cm\(^3\) each of solutions A and B in a test tube
(ii) adding excess dilute hydrochloric acid to the mixture from (a)(i) above
(b) 10cm\(^3\) of solution A were measured into each of seven boiling tubes of uniform bore and various quantities of solution B were added respectively to the boiling tubes, The tubes were immersed in hot water. After the reaction, the height of the product in each of the tubes was measured. The results were as tabulated below.
| Test tube | I | II | III | IV | V | VI | VII |
| Volume of sodium B added (cm\(^3\)) | 2.0 | 3.0 | 4.0 | 5.0 | 6.0 | 7.0 | 8.0 |
| Height of the product (to the nearest m) | 12 | 18 | 24 | 30 | 30 | 30 | 30 |
(i) Plot a graph of the height of the product against the volume of solution B added to each test tube. Explain the shape of the graph
(ii) Calculate the amount (in mole) of the trioxocarbonate (IV) salt contained in the volume of solution B added to tube lV
(iii) Calculate the amount (in mole) of barium chloride contained in 10cm\(^3\) of solution A. Hence, determine the mole ratio of barium chloride and the trioxocarbonate (IV) Salt in the reaction
Explanation
(a)(i) A white precipitate will be formed.
(ii) The precipitate will dissolve with effervescence. Confirmatory test of the gas evolved in (II): A colourless, odourless gas which turns lime water milky will be evolved. The gas is carbon (IV) oxide.
(b)(i).
The height of the precipitate increases as more of B is added until the reaction is complete in the test tube (IV). Thereafter the graph levels off because further addition of B produces no precipitation i.e. all the barium chloride has been used up.
(b)(ii) Volume of B added 5.0cm\(^3\);
Conc. of B = 10 mol dm See graph on above
Amount of B added (in mole) = \(\frac{5}{100}\) x 1.0 = 0.005 mole
(iii) Concentration of A = 0.5 mol dm\(^{-3}\)
Amount of A added in mole = \(\frac{10}{100} \times 0.5\) = 0.005 mole
Mole ratio of BaCl\(_2\); Trioxocarbonate (IV) salt
0.005 : 0.005 = 1 : 1