(a)(i) State two general methods of preparing soluble salts. (ii) Mention three pieces of apparatus...
(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