(a)(i) Explain what is meant by saturated solution
(ii) Describe in outline, a suitable procedure for preparing a saturated solution of sodium trioxonitrate(V) at 30°C.
(ii) State two techniques that can be used to recover crystals of sodium trioxonitrate(V) from its saturated solution.
(b) 1.0dm\(^3\) of an aqueous solution at 90°C contains 404g of potassium trioxonitrate(V) and 245g of potassium trioxochlorate (V).
(i) Determine which of the two salts will separate out when the solution is cooled to 60°C. N = 14. O = 16, CI = 35.5, K = 39; Solubility of KNO\(_3\) in water at 60\(^o\)C = 5.14 mol.dm\(^{-3}\), Solubility of KCIO\(_3\) in water at 60°C = 1.61 mol.dm\(^{-3}\)
(ii) Calculate the mass of salt that will separate out at 60°C
(c)(i) List two salts which cause hardness of water.
(ii) Explain why temporary hardness of water result in the furring of kettle.
(a)(i) A saturated solution is a solution that contains the maximum amount of dissolved solute it can hold at a given temperature, in the presence of (and in equilibrium with) undissolved solute.
(a)(ii) Preparing a saturated solution of \(NaNO_3\) at 30 C
- Warm some distilled water and keep it at 30 C in a thermostatic bath.
- Add sodium trioxonitrate(V) a little at a time, stirring, until no more will dissolve and some solid remains undissolved at 30 C.
- Filter off the excess undissolved solid; the filtrate is the saturated solution at 30 C.
(a)(iii) Two techniques to recover the crystals: evaporation of the solvent (evaporate to the point of crystallisation and allow to cool) and cooling/crystallisation of the hot saturated solution.
(b) Molar masses: \(KNO_3 = 39+14+48 = 101\ \text{g mol}^{-1}\); \(KClO_3 = 39+35.5+48 = 122.5\ \text{g mol}^{-1}\).
Amounts present in \(1.0\ dm^3\):
\[n(KNO_3) = \frac{404}{101} = 4.0\ \text{mol dm}^{-3}\]\[n(KClO_3) = \frac{245}{122.5} = 2.0\ \text{mol dm}^{-3}\]
(i) At 60 C the solubility of \(KNO_3\) is \(5.14\ \text{mol dm}^{-3}\); since \(4.0 < 5.14\), all the \(KNO_3\) stays in solution. The solubility of \(KClO_3\) is only \(1.61\ \text{mol dm}^{-3}\); since \(2.0 > 1.61\), the excess crystallises. Therefore potassium trioxochlorate(V), \(KClO_3\), separates out.
(ii) Amount separating \(= 2.0 - 1.61 = 0.39\ \text{mol}\).
\[\text{mass} = 0.39 \times 122.5 = 47.8\ \text{g}\]
(c)(i) Two salts causing hardness: calcium hydrogentrioxocarbonate(IV), \(Ca(HCO_3)_2\), and calcium tetraoxosulphate(VI), \(CaSO_4\) (magnesium salts also cause hardness).
(c)(ii) Temporary hardness furs a kettle because, on boiling, the dissolved \(Ca(HCO_3)_2\) decomposes to insoluble calcium trioxocarbonate(IV), which deposits as a hard scale (fur) on the kettle: \[Ca(HCO_3)_2 \xrightarrow{\text{heat}} CaCO_3 + H_2O + CO_2\]
(a)(i) A saturated solution is a solution that contains the maximum amount of dissolved solute it can hold at a given temperature, in the presence of (and in equilibrium with) undissolved solute.
(a)(ii) Preparing a saturated solution of \(NaNO_3\) at 30 C
- Warm some distilled water and keep it at 30 C in a thermostatic bath.
- Add sodium trioxonitrate(V) a little at a time, stirring, until no more will dissolve and some solid remains undissolved at 30 C.
- Filter off the excess undissolved solid; the filtrate is the saturated solution at 30 C.
(a)(iii) Two techniques to recover the crystals: evaporation of the solvent (evaporate to the point of crystallisation and allow to cool) and cooling/crystallisation of the hot saturated solution.
(b) Molar masses: \(KNO_3 = 39+14+48 = 101\ \text{g mol}^{-1}\); \(KClO_3 = 39+35.5+48 = 122.5\ \text{g mol}^{-1}\).
Amounts present in \(1.0\ dm^3\):
\[n(KNO_3) = \frac{404}{101} = 4.0\ \text{mol dm}^{-3}\]\[n(KClO_3) = \frac{245}{122.5} = 2.0\ \text{mol dm}^{-3}\]
(i) At 60 C the solubility of \(KNO_3\) is \(5.14\ \text{mol dm}^{-3}\); since \(4.0 < 5.14\), all the \(KNO_3\) stays in solution. The solubility of \(KClO_3\) is only \(1.61\ \text{mol dm}^{-3}\); since \(2.0 > 1.61\), the excess crystallises. Therefore potassium trioxochlorate(V), \(KClO_3\), separates out.
(ii) Amount separating \(= 2.0 - 1.61 = 0.39\ \text{mol}\).
\[\text{mass} = 0.39 \times 122.5 = 47.8\ \text{g}\]
(c)(i) Two salts causing hardness: calcium hydrogentrioxocarbonate(IV), \(Ca(HCO_3)_2\), and calcium tetraoxosulphate(VI), \(CaSO_4\) (magnesium salts also cause hardness).
(c)(ii) Temporary hardness furs a kettle because, on boiling, the dissolved \(Ca(HCO_3)_2\) decomposes to insoluble calcium trioxocarbonate(IV), which deposits as a hard scale (fur) on the kettle: \[Ca(HCO_3)_2 \xrightarrow{\text{heat}} CaCO_3 + H_2O + CO_2\]
