30 ml of a solution containing 9.15 gm/litre of an oxalate K_(x)H_(y) (C_(2)O_(4))_(z) nH_(2)O are rerquired for titrating 27 ml of 0.12 N NaOH and 36 ml of 0.12 N KMnO_(4) separately. Calculate x,y,zand n. Assume all H atoms (except H_(2)O) are replaceable and x,y,z are in the simple ratio of gm atoms.

30 ml of a solution containing 9.15 gm/litre of an oxalate K_(x)H_(y)

1.	30 ml of a solution containing 9.15 gm/litre of an oxalate K_(x)H_(y) (C_(2)O_(4))_(z) nH_(2)O are rerquired for titrating 27 ml of 0.12 N NaOH and 36 ml of 0.12 N KMnO_(4) separately. Calculate x,y,zand n. Assume all H atoms (except H_(2)O) are replaceable and x,y,z are in the simple ratio of gm atoms.
Answer» Solution :Let,molecular weight of oxalate salt is M (i) n-factor in ACID- base reaction = 2 (ii) n-factor in redox titration = `2 xx z` `(C_(2)O_(4)^(2-) to 2CO_(2) + 2e)` ` :. ` Meq. Of acid in 30 ml = Meq.of NaOH used `30 xx (9.15)/M xx y = 27 xx 0.12` Also, `30 xx (9.15)/M xx (2z) = 36 xx 0.12` From EQUATIONS (1) and (2)`y/(2z) = 27/36 rArry/z 3/2` Also, total cationic CHARGE = total anionic charge ` :. x + y = 2z` By equations (3) and (4) ` x : y : z :: 1 : 3 : 2` . These are in SIMPLEST ratio and molecular formula is `KH_(3)(C_(2)O_(4))_(2). nH_(2)O` Molecular weight of salt ` = 39 + 3 + 176 + 18n = 218 + 18n` Form equation (1), M ` = (30 xx 9.15 xx 3)/(27 xx 0.12) = 254. 16 ` ` :. 218 + 18n = 254.15` ` :. ` Oxalate salt is `KH_(3)(C_(2)O_(4))_(2) . 2H_(2)O`