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.

A sample of mixed alkalis containing NaOH and Na_(2)CO_(3) is titrated

1.	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.
Answer» Solution :Basic principle involved is as follows : Acid used with PHENOLPHTHALEIN as INDICATOR = Complete neutralization ofNaOH `+ (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 usedfor 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 ` 0.6meq 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`