(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})
Explanation
(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