'A' and 'B' are compounds of sodium, 'A' is thermally stable. On passing CO₂ through the solution of 'A', 'B' is formed. 'B' on heating gives 'A'. Phenolphthalein is added to the aqueous solution of 'A' and 'B'. The colours of solutions are

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Colouress, Colourless

Pink, Colourless

Colourless, Pink

Pink, pink

answer is B.

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Detailed Solution

mathop {N{a_2}C{O_3}}limits_{(A)} ; + ;{H_2}O; + ;C{O_2}; to ;mathop {2NaHC{O_3}}limits_{(B)} ;xrightarrow{{;;Delta ;;}};mathop {N{a_2}C{O_3}}limits_{(A)} ; + ;C{O_2}; + ;{H_2}O

aQ.Na₂CO₃ is more basic than aQ.NaHCO₃
Reason

large CO_3^{ - 2}; + ;2{H_2}O; to ;{H_2}C{O_3}; + ;2Omathop Hlimits^ ominus

large HCO_3^ - ; + ;{H_2}O; to ;{H_2}C{O_3}; + ;Omathop Hlimits^ominus

aQ.NaHCO₃ & Na₂CO₃ both alkaline gives yellow colour with basic indicator like methly orange
aQ.Na₂CO₃ being strong base gives pink colour with the acid indicator phenolyphthalein
aQ.NaHCO₃ is weakly alkaline and still contains acidic hydrogen and can't ionise the phenophthalein indicatior
Therefore, phenolphthalein indicator is colour less in aQ.NaHCO₃ solution, where as phenolphthalein indicator is becomes pink colour in aQ.Na₂CO₃ solution