(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:
(i) Plot the solubility curves of K and L on the same graph. Use the curves to answer questions (ii) - (iv) below.
(iii) At what temperature is the solubility of L equal to 1.0mol. dm\(^{-3}\)?
(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.
(a)(i) Two general methods of preparing soluble salts
- Neutralisation of a dilute acid with an alkali or a base (e.g. dilute acid + metal oxide/hydroxide).
- Action of a dilute acid on a suitable metal or on a metal trioxocarbonate(IV).
(a)(ii) Three pieces of apparatus for determining solubility at a given temperature
- Thermometer (to fix and read the temperature).
- Evaporating basin/dish (to evaporate a known volume of the saturated solution to dryness).
- Chemical (beam) balance (to weigh the residual salt and the basin).
(A beaker used as a water bath and a pipette are also acceptable.)
(b) The data
| 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 |
(b)(i) Plot solubility (y-axis) against temperature (x-axis) and draw a smooth curve through each set of points. K rises gently and almost linearly; L rises steeply, curving upward. The two curves cross near 31 \(^{o}\)C. The readings below are taken from these curves.
(b)(ii) Solubility of K at 50 \(^{o}\)C
50 \(^{o}\)C lies midway between 40 \(^{o}\)C (0.54) and 60 \(^{o}\)C (0.62):
\[ \text{Solubility} = \frac{0.54 + 0.62}{2} = 0.58\ \text{mol dm}^{-3} \]
(b)(iii) Temperature at which solubility of L = 1.0 mol dm\(^{-3}\)
1.0 lies between 40 \(^{o}\)C (0.64) and 60 \(^{o}\)C (1.08):
\[ t = 40 + \frac{1.00 - 0.64}{1.08 - 0.64}\times 20 = 40 + \frac{0.36}{0.44}\times 20 \approx 56\ ^{o}\text{C} \]
(b)(iv) Range over which K is more soluble than L
K exceeds L at 0 and 20 \(^{o}\)C but L overtakes K by 40 \(^{o}\)C. The curves cross where K = L:
\[ 0.46 + 0.004(t-20) = 0.34 + 0.015(t-20) \Rightarrow t \approx 31\ ^{o}\text{C} \]
So K is more soluble than L over the range 0 \(^{o}\)C to about 31 \(^{o}\)C.
(b)(v) Is 3.4 g of L per 250 cm\(^3\) at 20 \(^{o}\)C saturated or unsaturated? (molar mass of L = 101 g)
\[ \text{Amount of L} = \frac{3.4}{101} = 0.0337\ \text{mol} \]
\[ \text{Concentration} = \frac{0.0337\ \text{mol}}{0.250\ \text{dm}^3} = 0.135\ \text{mol dm}^{-3} \]
The solubility of L at 20 \(^{o}\)C (the concentration of the saturated solution) is 0.34 mol dm\(^{-3}\). Since 0.135 mol dm\(^{-3}\) is less than 0.34 mol dm\(^{-3}\), the solution can still dissolve more L, so it is unsaturated.
