In the mixture of `(NaHCO_(3)+Na_(2)CO_(3))` volume of HCl required is x mL with phenolphthalein indicator and then y mL with methyl orange indicator in same titration Hence, volume of HCl for complete reaction of `Na_(2)CO_(3)` is

Options

A. `2x`
B. `y`
C. `(x)/(2)`
D. `(y-x)`

Answer

D

Explanation

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`