Question

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 1

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`