0.2 gm of a mixture of NaOH and Na_(2)CO_(3) and inert impurities was first titrated with phenolphthalein and n/10HCl, 17.5 ml of HCl was required at the end point. After this methyl organge was added and 2.5 ml of same HCl was again required for next end point. Find out percentage of NaOH and Na_(2)CO_(3) in the mixture.

0.2 gm of a mixture of NaOH and Na_(2)CO_(3) and inert impurities was

1.	0.2 gm of a mixture of NaOH and Na_(2)CO_(3) and inert impurities was first titrated with phenolphthalein and n/10HCl, 17.5 ml of HCl was required at the end point. After this methyl organge was added and 2.5 ml of same HCl was again required for next end point. Find out percentage of NaOH and Na_(2)CO_(3) in the mixture.
Answer» Solution :Let , `W_(1)` gm NaOH and `W_(2) gmNa_(2)CO_(3)` was present in MIXTURE . At phenophthaiein end point , `(W_(1)/40 + 1/2xx W_(2)/53 ) = 1/10 xx 17.5 xx 10^(-3)` At SECOND end point following reaction takes place Eq. of `NaHCO_(3) = Eq`. of HCL used (in second titration ) ` = 1/2 " Eq. of " Na_(2)CO_(3)` ` 1/2 xx W_(2)/53 =25 xx 1/10 xx 10^(-3) ` `W_(2) = 0.0265` gm Putting the value of `W_(2) ` in Eq. (1) , we get `W_(1) = 0.06` gm Percentage of NaOH = `(0.06)/(0.2) xx 100 = 30 %` Percentage of `NaCO_(3) = (0.0265)/(0.2) xx 100 = 13.25 %`