(a)(i) State three methods of preparing salts, giving one example in each case of a salt so prepared.
(ii) What type of salt is each of the following? NaH\(_2\)PO\(_4\); (CH\(_3\)COO)\(_2\)Pb; KAI(SO\(_4\))\(_2\). 12H\(_2\)O.
(b)(i) Write an equation for the reaction between dilute HCI and a solution of AgNO\(_3\).
(ii) Explain why NaNO\(_3\) is preferred to AgNO\(_3\) in the preparation of oxygen by thermal decomposition of trioxonitrate (V) salts.
(iii) When silver wire was dipped into an aqueous solution of CuSO\(_4\), the wire remained intact but when the wire was replaced with zinc rod, the rod decreased in size. Give an explanation for this observation.
(c) When a sample of a crystalline salt X was exposed to air, there was a loss in mass.
(i) What phenomenon was exhibited by X?
(ii) Suggest two substances which X could be.
(iii) On heating 5.00 g of a fresh sample of X to constant mass, 1.80g was lost in the form of water vapour. Calculate the number of molecules of water of crystallization in one molecule of X. [H = 1.00; O = 16.00; Anhydrous form of X = 160 g mol\(^{-1}\)
(a)(i) Three methods of preparing salts
- Neutralisation of an acid by an alkali/base: \(HCl + NaOH \to NaCl + H_2O\) (gives sodium chloride).
- Action of an acid on a metal: \(Zn + H_2SO_4 \to ZnSO_4 + H_2\) (gives zinc tetraoxosulphate(VI)).
- Precipitation (double decomposition): \(AgNO_3 + NaCl \to AgCl + NaNO_3\) (gives silver chloride). Action of acid on a carbonate is also acceptable.
(a)(ii) Type of each salt
- \(NaH_2PO_4\): an acid salt (replaceable hydrogen remains).
- \((CH_3COO)_2Pb\): a normal salt (lead ethanoate).
- \(KAl(SO_4)_2\cdot12H_2O\): a double salt (potash alum).
(b)(i) \[HCl + AgNO_3 \to AgCl\downarrow + HNO_3\] (ionically \(Ag^+ + Cl^- \to AgCl\)).
(b)(ii) \(NaNO_3\) is preferred to \(AgNO_3\) because it decomposes simply to the nitrite and pure oxygen \((2NaNO_3 \to 2NaNO_2 + O_2)\), and it is cheap. \(AgNO_3\) decomposes to silver, oxygen and brown nitrogen(IV) oxide \((2AgNO_3 \to 2Ag + 2NO_2 + O_2)\), which contaminates the oxygen, and it is expensive.
(b)(iii) Silver is less reactive than copper, so it cannot displace copper and the wire stays intact. Zinc is more reactive than copper, so it displaces copper from the solution and itself dissolves, so the rod decreases in size: \[Zn + CuSO_4 \to ZnSO_4 + Cu\]
(c)(i) The phenomenon is efflorescence (loss of water of crystallisation to the air).
(c)(ii) Two possible substances: sodium trioxocarbonate(IV) decahydrate, \(Na_2CO_3\cdot10H_2O\), or sodium tetraoxosulphate(VI) decahydrate, \(Na_2SO_4\cdot10H_2O\).
(c)(iii) Mass of anhydrous salt \(= 5.00 - 1.80 = 3.20\ \text{g}\).
\[n(\text{anhydrous}) = \frac{3.20}{160} = 0.02\ \text{mol};\quad n(H_2O) = \frac{1.80}{18} = 0.10\ \text{mol}\]\[\frac{n(H_2O)}{n(\text{salt})} = \frac{0.10}{0.02} = 5\]
There are 5 molecules of water of crystallisation in one molecule of X.
(a)(i) Three methods of preparing salts
- Neutralisation of an acid by an alkali/base: \(HCl + NaOH \to NaCl + H_2O\) (gives sodium chloride).
- Action of an acid on a metal: \(Zn + H_2SO_4 \to ZnSO_4 + H_2\) (gives zinc tetraoxosulphate(VI)).
- Precipitation (double decomposition): \(AgNO_3 + NaCl \to AgCl + NaNO_3\) (gives silver chloride). Action of acid on a carbonate is also acceptable.
(a)(ii) Type of each salt
- \(NaH_2PO_4\): an acid salt (replaceable hydrogen remains).
- \((CH_3COO)_2Pb\): a normal salt (lead ethanoate).
- \(KAl(SO_4)_2\cdot12H_2O\): a double salt (potash alum).
(b)(i) \[HCl + AgNO_3 \to AgCl\downarrow + HNO_3\] (ionically \(Ag^+ + Cl^- \to AgCl\)).
(b)(ii) \(NaNO_3\) is preferred to \(AgNO_3\) because it decomposes simply to the nitrite and pure oxygen \((2NaNO_3 \to 2NaNO_2 + O_2)\), and it is cheap. \(AgNO_3\) decomposes to silver, oxygen and brown nitrogen(IV) oxide \((2AgNO_3 \to 2Ag + 2NO_2 + O_2)\), which contaminates the oxygen, and it is expensive.
(b)(iii) Silver is less reactive than copper, so it cannot displace copper and the wire stays intact. Zinc is more reactive than copper, so it displaces copper from the solution and itself dissolves, so the rod decreases in size: \[Zn + CuSO_4 \to ZnSO_4 + Cu\]
(c)(i) The phenomenon is efflorescence (loss of water of crystallisation to the air).
(c)(ii) Two possible substances: sodium trioxocarbonate(IV) decahydrate, \(Na_2CO_3\cdot10H_2O\), or sodium tetraoxosulphate(VI) decahydrate, \(Na_2SO_4\cdot10H_2O\).
(c)(iii) Mass of anhydrous salt \(= 5.00 - 1.80 = 3.20\ \text{g}\).
\[n(\text{anhydrous}) = \frac{3.20}{160} = 0.02\ \text{mol};\quad n(H_2O) = \frac{1.80}{18} = 0.10\ \text{mol}\]\[\frac{n(H_2O)}{n(\text{salt})} = \frac{0.10}{0.02} = 5\]
There are 5 molecules of water of crystallisation in one molecule of X.
