Given two mixtures: (I) NaOH and `Na_(2)CO_(3)` and (II) `NaHCO_(3)` and `Na_(2)CO_(3)`.
100 ml of 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

A-S; B-R; C-P; D-Q

Explanation

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`