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QUESTION
Q44) A solution containing Na2CO3 and NaOH requires 300 mL of 0.1 N HCl using phenolphthalein as an indicator. Methyl orange is then added to above titrated solution when a further 25 mL of 0.2 N HCl is required. The amount of NaOH present in solution is [ NaOH=40, Na2CO3=106] .. [ CPMT 1992]




SOLUTION

In presence of phenolphthalein, NaOH is completely neutralised and Na2CO3 is half neutralised (to NaHCO3). The remaining half is neutralised in the presence of methyl orange. Thus
For full neutralisation of NaOH + Half neutralisation of Na2CO3
0.1 N HCl required=300 mL
For half neutralisation of Na2CO3, HCl required=25 ml of 0.2N=50 ml of 0.1N
∴0.1N HCl required for complete neutralisation of NaOH=300 -50=250 mL
But 250 ml of 0.1N HCl=250 mL of 0.1N NaOH=1 g of NaOH(s)Answer: (b)