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30 " mL of " a solution containing 9.15g...

30 " mL of " a solution containing `9.15gL^(-1)`
of an oxalate `K_(x)H_(y)(C_2O_4)_2.nH_2O` required for titration 27 " mL of " 0.12 N NaOH and 36 " mL of " 0.12 N `KMnO_4` for oxidation Find x,y,z. and n.

Text Solution

Verified by Experts

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`
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