In the study of titration of NaOH and `Na_(2)CO_(3)`. NaOH and `NaHCO_(3)`, `Na_(2)CO_(3)` and `NaHCO_(3)`, phenophthalein and methyl orange are used as indicators.
(a). When phenolphthalein is used as an indicator for the above mixture:
(i). It indicates complete neutralisation of NaOH or KOH
(ii). It indicates half neutralisation of `Na_(2)CO_(3)` because `NaHCO_(3)` is formed at the end point.
(b). When methyl orange is used as an indicator for the above mixture
(i). It indicates complete neutralisation of `NaOH` or KOH
(ii). It indicates half neutralisation of `Na_(2)CO_(3)` because `NaCl` is formed at the end point.
Q. 1 L solution of `Na_(2)CO_(3)` and `NaOH` was made in `H_(2)O`. 100 " mL of " this solution required 20 " mL of " 0.4 M HCl in the presence of phenolphthalein however, another 100 mL sample of the same solution required 25 " mL of " the same acid in the presence of methyl orange as indicator. What is the molar ratio of `Na_(2)CO_(3)` and `NaOH` in the original mixture.

Options

A. `3:2`
B. `3:1`
C. `1:3`
D. `1:1`

Answer

C

Explanation

Let a and b m" Eq of "`Na_(2)CO_(3)` and NaOH are present in the mixture.
When phenolphtalein in used as indicator
`(a)/(2)m" Eq of "Na_(2)CO_(3)+bm" Eq of "NaOH=m" Eq of "HCl`
`(a)/(2)+b=20xx0.4xx1=8` .(i)
When methyl orange is used as indicator
`a m" Eq of "Na_(2)CO_(3)+b m" Eq of "NaOH=m" Eq of "HCl` ltbr `a+b=25xx0.4xx1=10` ..(ii)
Solve for a and b from equation (i) and (ii)
`a=4 mEq=(4)/(2)mmol=2mmol`
`b=6mEq=(6)/(1)mmol=6mmol`