Solution ''X'' contains `Na_(2)CO_(3)` and `NaHCO_(3),  20 ml` of X when titrated using methyl organge indicator consumed 60 ml of 0.1M HCI solution. In another experiment, 20 ml of X solution when titrated using phenolphthalein consumed 20 ml of `0.1 M HCl` solution. The concentraions `("in mol lit"^(-1))` of `Na_(2)CO_(3) and NaHCO_(3)` in X are respectively)

Options

A. 0.01,0.02
B. 0.1,0.1
C. 0.01,0.01
D. 0.1,0.01

Answer

(b)

Explanation

(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