Determination of Molecular Mass of Base by Chloroplatinate Method

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° = √W_salt/√W_Pt = √3
or, tan 60° = W_salt/W_Pt = 3
B₂H₂PtCl₆ → Pt
W_saltW_Pt
POAC on Pt
[W_salt/(M x 2 + 410 x 4)] x 4 = W_Pt/195
W_salt/W_Pt) x 4 x 195 = 2M − 1640
3 x 4 x 195 = 2M − 1640
2340 − 1640 = 2M
or, M = 700/2
M = 350.

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