(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) - By the action of a dilute acid on a metal e.g. Zn + 2HCI --> ZnCl\(_2\) + H\(_2\) or Mg + H\(_2\)SO\(_4\) \(\to\) MgSO\(_4\) + H\(_2\)
- By neutralisation of an alkali by an acid e.g NaOH + HCI -> NaCI + H\(_2\)O
-By the action of a dilute acid on an insoluble base e.g 2HCI + MgO -> MgCl\(_2\) + H\(_2\)O
-By the action of a dilute acid on a trioxocarbonate (IV) e.g 2HCI + CaCO\(_2\) \(\to\) CaCl\(_2\) ± H\(_2\) O + CO\(_2\)
(ii) NaH\(_2\)PO\(_2\) -An acid salt;
(CH\(_3\)COO)\(_2\) Pb -An organic salt
KAI(SO\(_4\))\(_2\) . 12H\(_2\)O - hydrated double salt
(b)(i) HCI + Ag NO\(_3\) -> AgCI \(\downarrow\) + HNO\(_3\)O
(ii) NaNO\(_3\) is preferred to AgNO\(_3\) in the thermal decomposition to give O\(_2\) because NaNO\(_3\) decomposes to only solid and O\(_2\) gas making it easy to collect, unlike AgNO\(_3\) which decomposes into silver, oxygen and nitrogen (IV) oxide.
(iii) Silver cannot displace Cu\(^{2+}\) in the solution of CuSO\(_4) but Zn can, because Zn is higher than Cu in the activity series but Ag is lower than Cu.
(c)(i) Efflorescence
(ii) Na\(_2\)CO\(_3\). 10H\(_2\)O or CuSO\(_4\) .10H\(_2\)O.
-mH\(_2\)O
(iii) X .mH\(_2\)O \(\to\) X. from the heat equation 160 + 18m --> 160 ; 5 ->1.80
heat
\(\frac{160 + 18m}{5}\) = \(\frac{160}{1.80}\)
288 + 32.4m = 512
m = \(\frac{512}{32.4}\)
= 15.8
m = 16 molecules
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