Correct Answer – A
In pressure of phenolphthalein, `50% Na_(2)CO_(3)` is neutralised whereas `NaHCO_(3)` remains unaffected. In presence of methyl orange, both `Na_(2)CO_(3) ” and” NaHCO_(3)` will be 100% neutralised.
Let volume of HCl for complete reaction of `Na_(2)CO_(3)=V_(1)` mL and volume of HCl for complete reaction of `NaHCO_(3)=V_(2)` mL . With phenolphthalein, `50% Na_(2)CO_(3)` will be neutralized.
`therefore ” ” (V_(1))/(2)=x, V_(1)=2x`

Correct Answer – D
With phenolphthalein indicator: `NaHCO_(3)` does not react with HCl whereas `Na_(2)CO_(#)` reacts upto `NaHCO_(3)` stage `(50%` reaction).
`V_(HCl)=x mL`
(ii). With methyl orange indicator : `NaHCO_(3)` reacts completely with HCl and with `Na_(2)CO_(3)` is `100%` reaction. But y ” mL of ” HCl is added after `Na_(2)CO_(3)` has reacted upto `NaHCO_(3)`. (i.e., half titre value of `Na_(2)CO_(3)`)
`V_(HCl)=` Full titre value of `NaHCO_(3)+` half titre value of `Na_(2)CO_(3)`.
`ymL=` full titre value of `NaHCO_(3)+x mL`
Full titre value of `NaHCO_(3)=(y-x)mL`

Correct Answer – D