How many " mL of " 0.1 N HCl is required to react completely using phenolphthalein with 2.0 g mixture of `Na_2CO_3` and `NaOH` containing equimolar amounts of two?

How many " mL of " 0.1 N HCl is required to react completely using phe

1.	How many " mL of " 0.1 N HCl is required to react completely using phenolphthalein with 2.0 g mixture of `Na_2CO_3` and `NaOH` containing equimolar amounts of two?
Answer» Let x mol each of `Na_2CO_3` and `NaOH` be there in `2.0g` Mixture. So `x(106)+x(40)=2impliesx=(1)/(73)` In neutralisation: g " Eq of "`Na_2CO_3=2x` But only half will be neutralised in phenolpthalein. g " Eq of "`NaOH=x`, fully neutralised `implies(1)/(2)(2x)+x=g m" Eq of "HCl=(1)/(10)V` `implies2x=(V)/(10)` or `V=20x=20((1)/(73))=0.274L-=274mL`

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How many " mL of " 0.1 N HCl is required to react completely using phenolphthalein with 2.0 g mixture of `Na_2CO_3` and `NaOH` containing equimolar amounts of two?

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