(a)(i) State two general methods of preparing soluble salts.
(ii) Mention three pieces of apparatus required for determining the solubility of a salt at a given temperature.
(b) The solubilities of two salts represented as K and L were determined at various temperatures. The results are shown in the table below:
| Temperature ((^o)C) | 0 | 20 | 40 | 60 | 80 | 90 |
| Solubility of K (mol. dm(^{-3})) | 0.38 | 0.46 | 0.54 | 0.62 | 0.69 | 0.73 |
| Solubility of L (mol. dm(^{-3})) | 0.12 | 0.34 | 0.64 | 1.08 | 1.64 | 2.00 |
(i) Plot the solubility curves of K and L on the same graph. Use the curves to answer questions (ii) – (iv) below.
(ii) What is the solubility of K at 50°C?
(iii) At what temperature is the solubility of L equal to 1.0mol. dm(^{-3})?
(iv) Over what temperature range is K more soluble that L?
(v) Given that the molar mass of L is 101g, determine whether a solution containing 3.4g of L per 250cm(^3) at 20°C is saturated or unsaturated.
Explanation
(a)(i) – Neutralization of a base by an acid
– Direct combination of elements
(ii) Thermometer, water bath and stirrer
(b)(i) The solubility of K at 50(^o) equal to 0.58mol/dm(^3)
(ii) The solubility of L equals to 1.0mol/dm(^{-3}) at 57(^o)C
(iii) K is more soluble than L over the temperature range of 0(^o) to 31.5(^o)C
(iv) Molar mass of L = 101g
3.4gm of L per 250cm(^3) (to) 3.4 (to) 250X (to) 1000X = (frac{1000}{250}) x 3.4 = 13.6g/dm(^3)
At 20(^o)C the molar conc. of L = 0.34
The solubility 101 x 0.34 = 34.34g/dm(^3) at 20(^o)C. Since the concentration/solubility of 13.6g/dm(^3) is lower than the required solubility of 34.34g/dm(^3) at 20(^o)C, it is therefore not saturated ie. unsaturated