6.4 g of a pure monobasic organic acid is burnt completely in excess of oxygen and CO_(2) evolved is absorbed completely in one litre of an aqueous solution of NaOH. A 10 mL portion of this solution required 14.5 mL of a normal HCl solution to reach the phenolphthalein end point. An another 10 mL portion of the same solution required 18 mL of the same HCl solution to reach the methyl organe end point. If the organic acid contains 25% oxygen by weight, deduce the empirical formula of this acid and strength of original NaOH solution.

6.4 g of a pure monobasic organic acid is burnt completely in excess o

1.	6.4 g of a pure monobasic organic acid is burnt completely in excess of oxygen and CO_(2) evolved is absorbed completely in one litre of an aqueous solution of NaOH. A 10 mL portion of this solution required 14.5 mL of a normal HCl solution to reach the phenolphthalein end point. An another 10 mL portion of the same solution required 18 mL of the same HCl solution to reach the methyl organe end point. If the organic acid contains 25% oxygen by weight, deduce the empirical formula of this acid and strength of original NaOH solution.
Answer» Solution :`C_(x) H_(y) O_(2)+O_(2) rarr x CO_(2) overset("2 NaOH")(rarr) xNa_(2) CO_(3)` ` ( 6.4)/( M)``( 6.4x)/( M) ``( 6.4x)/( M)` Let 100ml of solution contains a MEQ. Of NaOH & b meq. Of `Na_(2) CO_(3)` when `HPh` is used as indicator then meq. of NaOH `+ 1//2` meq. of `Na_(2) CO_(3) -` Meq. of HCl, `= ( 14.5xx 1)/( 10) xx 1000 = 40%` `a + b//2 = 14.5 xx 100 = 1450`....(1) when MeOH is used as indicator, meq. of `NaOH + `meq. of `Na_(2) CO_(3)` = Meq. of HCl `a + b = 18 xx 1 xx(1)/( 10) xx 100 = 1800 `.....(2) from ( 1) & ( 2) `b = 700,a = 100,` moles of `Na_(2) CO_(3) = ( b xx 10^(-3))/( 2) = ( 6.4) /(M) xx x ` `rArr( 700 xx 10^(-3))/( 2) = 6.4 ( x )/(M).....(1) ` , wt% of =25 `M xx ( 25)/( 100) = 32rArr M = 128` from ( 1) `( 700 xx 10^(-3))/( 2) = 6.4 xx ( x)/( 128)` `( 700 xx 10^(-3))/( 2)xx ( 128)/( 6.4) = x = 7` formula `C_(7) H_(12) O_(2)` Molarity of NaOH `= ( 1800)/( 1000) = 1.8 M`