Determination of Molecular Weight of a Base: Chloroplatinate (H2PtCl6) Method
Most of the bases combine with hydrochloroplatinic acid to form insoluble salts known as platinichlorides. These double salts may be represented by the general formula B2H2PtCl6 where B2 stands for one equivalent of the base.
On ignition, H2PtCl6 decomposes to leave a residue of Pt.
B2H2PtCl6 → Pt
Knowing the weight of B2H2PtCl6 taken and the Pt left as residue, the molecular weight of the base can be calculated.
Let 'x' gm be the weight of the B2H2PtCl6 taken and 'a' of the Pt residue.
Since one molecule of B2H2PtCl6 contains one Pt atom. So, 195 gm Pt will be left as residue by one molecular weight of the H2PtCl6.
But, a gm of Pt left by x gm of B2H2PtCl6
So, 195 gm of Pt will left by (x/a)195 gm
Hence, molecular weight of B2H2PtCl6 = (x/a)195
The equivalent weight of the base (B) = (B2H2PtCl6 − H2PtCl6)/2
or, (mol. wt of H2PtCl6 − 410)/2
or, [(x/a)195 − 410]/2
If the acidity of the base is 'n', the molecular weight of the base-
To find molecular mass of base (acidity =4) by chloroplatinate salt method following graph is plotted. Molecular weight of base will be where y = root of weight of platinum salt, x = root of weight of residue.
Answer
tan 60° = √Wsalt/√WPt = √3
or, tan 60° = Wsalt/WPt = 3
B2H2PtCl6 → Pt
WsaltWPt
POAC on Pt
[Wsalt/(M x 2 + 410 x 4)] x 4 = WPt/195
Wsalt/WPt) x 4 x 195 = 2M − 1640
3 x 4 x 195 = 2M − 1640
2340 − 1640 = 2M
or, M = 700/2
M = 350.