Match the statements given in Column I abd Column II.

Answer

`A to (r, s), B to (q, t) , C to (p) D to (s, t)`

Explanation

(A) `Na_(2)CO_(3) + NaHCO_(3)`
Using phenolphthalein indicator .
eq of `Na_(2)CO_(3)` will react.
` :. 1^(st)` titre value
`Na_(2) CO_(3) + HCl to NaHCO_(3)+ NaCl`
` :. 1^(st) ` titre value = eq of `Na_(2)CO_(3)`
Using methyl orange indicator :
eq of `NaHCO_(3) ` produced from `Na_(2)CO_(3)` & eq of `NaHCO_(3)` present will react :
`NaHCO_(3) + HCl to NaCl + H_(2)O + CO_(2)`
` :. 2^(nd) ` titre value = eq of `Na_(2)CO_(3) + " eq of " NaHCO_(3)` .
` A to r, s`
(B) `Na_(2)CO_(3) + NaOH`
Using phenolphthalein indicator :
`Na_(2)CO_(3)+HCl to NaHCO_(3) + NaCl`
` NaOH + HCl to NaCl + H_(2)O`
` :. 1^(st) ` titre value = eq of `Na_(2)CO_(3) + ` eq of NaOH .
Using methyl orange indicator :
`NaHCO_(3) + HCl to NaCl + H_(2)O + CO_(2)`
(produced from `Na_(2)CO_(3)`)
`2^(nd) ` titre value = eq of `Na_(2)CO_(3)`
` B to q,t `
(C ) `Na_(2)C_(2)O_(4) + H_(2)C_(2)O_(4)`
Using `NaOH ` & phenolphthalein indicator :
` H_(2)C_(2)O_(4) + NaOH to Na_(2)C_(2)O_(4) + H_(2)O`
` :. 1^(st)` titre value of NaOH = eq. of `H_(2)C_(2)O_(4)`
Using `KMnO_(4)` in acidic medium :
Using `KMnO_(4)` in acidic medium :
`KMnO_(4) + Na_(2)C_(2)O_(4) + H^(+) to CO_(2) + Mn^(2+) + H_(2)O + Na^(+)`
`2^(nd)` titre value will give the eq of total `Na_(2)C_(2)O_(4) & H_(2)C_(2)O_(4)` but in different n-factor value .
The estimation of `Na_(2)C_(2)O_(4)` can be done by the difference in moles of total `Na_(2)C_(2)O_(4) + H_(2)C_(2)O_(4) " in " 2^(nd) ` titration & moles of `H_(2)C_(2)O_(4) " in " 1^(st)` titration .
But difference in `1^(st) & 2^(nd)` titre value will not give the estimation of `1^(st)` constituent.
` C to P `
(D) `H_(2)CO_(3) + NaHCO_(3)`
Using NaOH & methyl orange indicator :
`H_(2)CO_(3) + NaOH to NaHCO_(3) + H_(2)O`
`1^(st)` titre value will give the estimation of `1^(st)` constituent.
Using `NaOH` & phenolphthalein indicator :
`NaHCO_(3) + NaOH to Na_(2)CO_(3) + H_(2)O`
`uparrow`
`(NaHCO_(3) " present + " NaHCO_(3) " preduced from " H_(2)CO_(3))`
Difference in `2^(nd) & 1^(st)` titre value will give the estimation of `2^(nd)` constituent .
`D to s, t`