Question

1.0 g sample of the Rochelle salt
`[NaOOC-CHOH-CHOH-COOK]`
`(NaKC_(4)H_(4)O_(6).4H_(2)O)(Mw=282)`, on ignition, is converted
into `NaKCO_(3)(Mw=122)`, which is titrated with 50 " mL of " 0.1 `MH_(2)SO_(4)`. The excess of `H_(2)SO_(4)` requries 30 mL 0.2 M KOH.

Answer 1

`NaKC_(4)H_(4)O_(6).4H_(2)Ooverset(Delta)toNaK.CO_(3)+3CO_(2)uarr+H_(2)Ouarr`
Total m" Eq of "`H_(2)SO_(4)=50xx0.1xx2` (n factor) `=10.0`
Excess m" Eq of "`H_(2)SO_(4)=m" Eq of "NaOH used`
`=30xx0.2xx1` (n factor)
`=6.0`
m" Eq of "`H_(2)SO_(4)` used for `NaKCO_(3)`
`=m" Eq of "H_(2)SO_(4)` (total)`-m" Eq of "H_(2)SO_(4)` used for NaOH
`=10-6=4.0`
`thereforem" Eq of "NaKCO_(3)=4.0`
`implies` n-factor of `NaKCO_(3)=2`, during its neutralisation with
`H_(2)SO_(4)(" Eq of "NaKCO_(3)=(122)/(2))`
Weight of `NaKCO_(3)=4.0xx10^(-3)xx(122)/(2)=0.244g`
1 " mol of "Rochelle salt`=1 " mol of "NaKCO_(3)`
Therefore, weight of rochelle salt`=(282xx0.244)/(122)=0.364g`
`%` purity of Rochelle salt`=(0.364xx100)/(1.0)=36.4%`