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NCERT Class XI Chemistry The p-Block Elements Solutions

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Question : 30 of 38
Marks: +1, -0
A certain salt X, gives the following results :
(i) Its aqueous solution is alkaline to litmus.
(ii) It swells up to a glassy material Y on strong heating.
(iii) When conc. H2SO4\mathrm{H_2SO_4} is added to a hot solution of X, white crystal of acid Z separates out.
Write equations for all the above reactions and identify X, Y and Z.
Solution:  
The data suggests that the salt ‘X’ is Borax (Na2B4O7⋅10H2O)\mathrm{(Na_2B_4O_7\cdot10H_2O)}.
(i) The aqueous solution of borax is basic in nature due to hydrolysis and turns red litmus blue.
Na2B4O7(X)+7H2O\underset{(X)}{\mathrm{Na_2B_4O_7}}+\mathrm{7H_2O} → 4H3BO3(Weak acid)+2NaOH(Strong base)\underset{\text{(Weak acid)}}{\mathrm{4H_3BO_3}}+\underset{\text{(Strong base)}}{\mathrm{2NaOH}}
(ii) Borax swells in size upon strong heating and loses molecules of water of crystallisation to form solid (Y).
Na2B4O7⋅10H2O\mathrm{Na_2B_4O_7 \cdot 10H_2O} →−10H2Oheat\xrightarrow[-\mathrm{10H_2O}]{\text{heat}} Na2B4O7\mathrm{Na_2B_4O_7} →strong heating\xrightarrow[]{{\text{strong heating}}} 2NaBO2+B2O3Glassy (y)\underset{\text{Glassy (y)}}{\mathrm{2NaBO_2+B_2O_3}}
(iii) Upon reacting with conc. H2SO4\mathrm{H_2SO_4}, borax forms boric acid (H3BO3\mathrm{H_3BO_3}).
When crystallised from the solution, it is in the form of white crystals (Z).
Na2B4O7+H2SO4+5H2O\mathrm{Na_2B_4O_7+H_2SO_4+5H_2O} → Na2SO4+4H3BO3(white crystals)(Z)\underset{\text{(white crystals)(Z)}}{\mathrm{Na_2SO_4+4H_3BO_3}}
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