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

In the mixture of (NaHCO_(3) + Na_(2)CO_(3)), volume of HCl required i

1.	In the mixture of (NaHCO_(3) + Na_(2)CO_(3)), volume of HCl required is x mL with phenolphthalein indicator and y mL with mthyl orange inidicator in same titration. Hence, volume of HCl for complete reaction of Na_(2)CO_(3) is :
Answer» 2 x y `x//2` `(y-x)` Solution :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`