40 mL of 0.05 M solution of sodium sesquicarbionate (Na_(2)CO_(3). NaHCO_(3).2H_(2)O) is titrated against 0.05 M HCl solution. X ml of HCl solution is used when phenophthalein is the indicator and y ml of HCl is used when methyl orange is the indicator in two separate titrations Hence (y-x) is

40 mL of 0.05 M solution of sodium sesquicarbionate (Na_(2)CO_(3). NaH

1.	40 mL of 0.05 M solution of sodium sesquicarbionate (Na_(2)CO_(3). NaHCO_(3).2H_(2)O) is titrated against 0.05 M HCl solution. X ml of HCl solution is used when phenophthalein is the indicator and y ml of HCl is used when methyl orange is the indicator in two separate titrations Hence (y-x) is
Answer» 80 mL 30 mL 120 mL none of these Solution :In the first TITRATION EQ. HCL = `(1)/(2)` eq `NC_(2)CO_(3)` `(x xx0.05)/(1000)=(1)/(2)XX(0.050xx40xx2)/(1000)impliesx=40ml` In the second titration Eq of HCl = Eq `NC_(2)CO_(3)+` Eq `NKHCO_(3)` `(yxx0.05)/(1000)=(40xx0.05xx2)/(1000)+(40xx0.05xx1)/(1000)` `y=120impliesy-x=80`