# Determination of Molecular Weight of a Base: Chloroplatinate (H_{2}PtCl_{6}) 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 B_{2}H_{2}PtCl_{6} where B_{2} stands for one equivalent of the base.

On ignition, H_{2}PtCl_{6} decomposes to leave a residue of Pt.

B_{2}H_{2}PtCl_{6} → Pt

Knowing the weight of B_{2}H_{2}PtCl_{6} taken and the Pt left as residue, the molecular weight of the base can be calculated.

Let 'x' gm be the weight of the B_{2}H_{2}PtCl_{6} taken and 'a' of the Pt residue.

Since one molecule of B_{2}H_{2}PtCl_{6} contains one Pt atom. So, 195 gm Pt will be left as residue by one molecular weight of the H_{2}PtCl_{6}.

But, a gm of Pt left by x gm of B_{2}H_{2}PtCl_{6}

So, 195 gm of Pt will left by (x/a)195 gm

Hence, molecular weight of B_{2}H_{2}PtCl_{6} = (x/a)195

The equivalent weight of the base (B) = (B_{2}H_{2}PtCl_{6} − H_{2}PtCl_{6})/2

or, (mol. wt of H_{2}PtCl_{6} − 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

B_{2}H_{2}PtCl_{6} → Pt

W_{salt}W_{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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