(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 g...
Question 1 Report
(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 \( (^{o}\mathrm{C}) \) | 0 | 20 | 40 | 60 | 80 | 90 |
| Solubility of K \( (\mathrm{mol.\ dm}^{-3}) \) | 0.38 | 0.46 | 0.54 | 0.62 | 0.69 | 0.73 |
| Solubility of L \( (\mathrm{mol.\ dm}^{-3}) \) | 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 \(1.0\mathrm{mol.\ dm}^{-3}\)?
(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 \(250\mathrm{cm}^{3}\) at 20°C is saturated or unsaturated.
Answer Details
(a)(i) General methods of preparing soluble salts
- Neutralisation of an acid by an alkali or base.
- Direct combination of the constituent elements.
(a)(ii) Apparatus required
A thermometer, a water bath and a glass stirrer.
(b)(i) Solubility curves
The solubility curves of K and L are plotted below on the same axes.
(b)(ii) From the curve for K, at 50°C:
Solubility of K = 0.58 mol dm−3.
(b)(iii) From the curve for L, a solubility of 1.0 mol dm−3 is obtained at approximately 57°C.
(b)(iv) K is more soluble than L from 0°C to about 31.5°C.
(b)(v)
Mass concentration of L in the solution is:
\[\frac{3.4\ \text{g}}{250\ \text{cm}^3}\times 1000 = 13.6\ \text{g dm}^{-3}\]
At 20°C, the solubility of L is 0.34 mol dm−3. In mass units:
\[0.34\times 101 = 34.34\ \text{g dm}^{-3}\]
Since 13.6 g dm−3 is less than 34.34 g dm−3, the solution is unsaturated.
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