Solution "X" contains Na_(2)CO_(3) and NaHCO_(3) 20 mL of X when titrated using methyl orange indicator consumed 60 mL of 0.1 M of X solution, when titrated using phenolphthalein consumed 20 ml. of 0.1 M HCI solution. The concentrations (in mol L^(-1) ) of Na_(2)CO_(3) and NaHCO_(3), in X are respectively

Solution "X" contains Na_(2)CO_(3) and NaHCO_(3) 20 mL of X when titra

1.	Solution "X" contains Na_(2)CO_(3) and NaHCO_(3) 20 mL of X when titrated using methyl orange indicator consumed 60 mL of 0.1 M of X solution, when titrated using phenolphthalein consumed 20 ml. of 0.1 M HCI solution. The concentrations (in mol L^(-1) ) of Na_(2)CO_(3) and NaHCO_(3), in X are respectively
Answer» 0.01,0.02 0.1,0.1 0.01,0.01 0.1,0.01 Solution :(i) If phenolphthalein is used as indicator for titration of `Na_(2)CO_(3)` with HCl ,the END point is reached only when half of the `Na_(2)CO_(3)` neutralised. `Na_(2)CO_(3) + HCI rarr NAHCO_(3) + NaCl` At the end point, there is no reaction between `NaHCO_(3)` and HCI. Volume of `Na_(2)CO_(3)` = Volume of HCl CONSUMED 20mL of 0.1 M=20mL of 0.1 M `:.`The concentration of `Na_(1)CO_(3)` in given solution 0.1M Since only half volume of HCl is consumed for titration of `Na_(2)CO_(3)` at phenolphthalein end point. The remaining half volume of HCI is used for complete NEUTRALIZATION of `Na_(2)CO_(3)` with methyl orange as indicator `NAHCO_(3) + HCI rarr NaCI + CO_(2)+ H_(2)O` `:.`Volume of HCI used for complete neutralization of `Na_(2)CO_(3)` in given solution = `2 xx20 ml = 40 ml` (ii) The remaining HCl used to neutralize `NaHCO_(3)` from the given X solution = 60– 40= 20 ml of 0.1 M From the equation Volume of `NaHCO_(3)`= Volume of HCl consumed 20 mL of 0.1M = 20 mL of 0.1 M `:.`The concentration of `NaHCO_(3)`in given X solution = 0.1M