A sample of mixed alkalis containing NaOH and `Na_(2)CO_(3)` is titrated in the following two schemes :
(i) 10 ml of above mixture requires 8 ml of 0.1 N HCl by using phenolphthalein.
(ii) 10 ml of above mixture requires 10 ml of 0.1 N HCl by using methyl orange.
Calculate the ratio of the weight of NaOH and `Na_(2)CO_(3)` in the sample mixture.

Explanation

Basic principle involved is as follows :
Acid used with phenolphthalein as indicator = Complete neutralization of NaOH `+ (1)/(2)` neutralization of `Na_(2)CO_(3)`
Acid used with methyl orange as indicator = Complete neutralization of NaOH + complete neutralization of `Na_(2)CO_(3)`
meq of HCl used for 10 ml of mixture using phenolphthalein `= 8 xx 0.1 = 0.8`
`:. ` meq of `NaOH + (1)/(2) M "eq of" Na_(2)CO_(3) = 0.8` ...(i)
meq of HCl used for 10 ml of mixture using methly orange `= 10 xx 0.1 = 1 `
`:.` meq of NaOH + M eq of `Na_(2)CO_(3)=1`
From eqns (i) and (ii), meq of `Na_(2)CO_(3) = (1-0.8)xx2=0.4 ,` meq of `NaOH = 1 - 0.4 = 0.6`
Eq. wt of `Na_(2)CO_(3) = 106//2=53`, Eq. wt of NaOH = 40
`:. ` 0.4 meq of `Na_(2)CO_(3) = (53)/(1000) xx0.4 = 0.0212 g `
0.6 meq of `NaOH = (40)/(1000) xx 0.6 = 0.024 g`
Ratio of weight of NaOH and `Na_(2)CO_(3)=0.024//0.0212 = 1.132`