Given two mixtures: (I) NaOH and Na_(2)CO_(3) and (II) NaHCO_(3) and Na_(2)CO_(3). 100 mlof mixture I required w and x ml of 1 M HCl is separate titrations using phenophthalein and methyl orange indicators while 100 ml of mixture II required y and z ml of same HCl solutioni in separate titrations using the same indicators.

Given two mixtures: (I) NaOH and Na_(2)CO_(3) and (II) NaHCO_(3) and N

1.	Given two mixtures: (I) NaOH and Na_(2)CO_(3) and (II) NaHCO_(3) and Na_(2)CO_(3). 100 mlof mixture I required w and x ml of 1 M HCl is separate titrations using phenophthalein and methyl orange indicators while 100 ml of mixture II required y and z ml of same HCl solutioni in separate titrations using the same indicators.
Answer» Solution :If Mixture I has NaOH = a m.moles `Na_(2)CO_(3)` = b m. moles and mixture II has `NaHCO_(3)` = C m.moles `Na_(2)CO_(3)` = d m.moles Mixture-I titration : Phenolphthalein : EQ HCl = eq. `NaOH+(1)/(2)` eq.`Na_(2)CO_(3)` `W=a+b` Methyl orange : eq. HCl = eq. NaOH + eq. `Na_(2)CO_(3)` `x=a+2b` `impliesa=2w-ximplies[NaOH]=(2w-x)/(100)M` `b=x-wimplies[Na_(2)CO_(3)]=(x-w)/(100)M` Mixture II Titration : Phenolphthalein : eq. `HCl=(1)/(2)` eq. `Na_(2)CO_(3)` `y=d` Methyl orange : eq. HCl = eq. `NaHCO_(3)+` eq. `Na_(2)CO_(3)` `z=c+2d` `IMPLIES d=yimplies[Na_(2)CO_(3)]=(y)/(100)M` `c=z-2dimplies[NaHCO_(3)]=(z-2y)/(100)M`