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 the above titrated solution when a further $$25$$ ml of $$0.2$$ N HCl is required. The amount of NaOH present in solution is $$NaOH=40$$, $$Na2CO3=106$$

Question

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 the above titrated solution when a further $$25$$ ml of $$0.2$$ N HCl is required. The amount of NaOH present in solution is $$NaOH=40$$,  $$Na2CO3=106$$

Infinity Learn · Accepted Answer

(I)  Phenolphthalein indicate partial neutralization of $$Na2CO3$$ →  $$NaHCO3Meq$$.  of $$Na2CO3$$  $$+$$  Meq. of  $$NaOH=Meq$$.  of  HClWE ×  $$1000+WE$$  ×  $$1000=NVSuppose$$  $$Na2CO3=a$$ g,  $$NaOH=b$$ $$ga53$$  ×  $$1000+b40$$ ×  $$1000$$ $$=$$ $$300$$ ×  $$0.1$$      .....$$.1(II)$$ Methyl orange indicates  complete neutralization    HCl             HCl     $$N1V1=N2V2$$,   $$25$$  ×  $$0.2=0.1$$  ×  $$V2$$  so  $$V2=50$$  ml  excess∴    $$a53$$  ×  $$1000$$  $$+$$ $$b40$$ ×  $$1000=350$$ ×  $$0.1$$       ....$$.2From$$   $$1$$  and $$2$$  $$b=1$$  gm.

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 the 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

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