1.

State one relevant observation for each of the following reactions : (i) Lead nitrate solution is treated with sodium hydroxide solution drop wise till it is in excess. (ii) At the anode, when molten lead bromide is electrolyzed using graphite electrodes. (iii) Lead nitrate solution is mixed with dilute hydrochloric acid and heated. (iv) Anhydrous calcium chloride is exposed to air for some time. (v) Barium chloride solution is slowly added to sodium sulphate solution.

Answer»

(i) On dropwise addition of sodium hydroxide solution to lead nitrate solution it first gives a white precipitate and then on adding excess of sodium hydroxide solution, a clear solution is obtained due to formation of sodium plumbate (Na2PbO2) which is colourless and soluble.

Pb(NO3)2 + 2NaOH ---> Pb(OH)2 (↓) + 2NaNO3 

Pb(OH)2 + 2NaOH (Excess) ---> Na2PbO2 

(ii) At the anode, when lead bromide is electrolyzed using graphite electrodes following reaction occurs at the anode during electrolysis :

2Br- + 2e- ---> Br2(↑) 

(iii) Lead nitrate solution is mixed with dilute hydrochloric acid and heated to give lead chloride and nitric acid : 

Pb(NO3)2 + 2HCl --> PbCl2 + 2HNO3 

(iv) Anhydrous calcium chloride is exposed to air for some time and it absorbs moisture from air as it has a strong affinity for water : 

CaCI2 + 2H2O ---> Ca(OH)2 +2HCl

(v) Barium chloride solution is slowly added to sodium sulphate solution to obtain barium sulphate : 

BaCl2(aq) + Na2SO4(aq) ---> BaSO4(s) + 2NaCl(aq)


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