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,z and n. Assume all H atoms (except `H_(2)O`) are replaceable and x,y,z are in the simple ratio of gm atoms.

Explanation

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 rArr y/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`