Question

A saturated of K2PtCl contains 11 mg of salt in each mL.
Ar K: 39.1
Ar Cl: 35
Ar Pt: 195
1.what is the solubility product of the salt

Answer 1

$No of moles of K_2PtCl_6 = 11/483.2 ×10^{-3} = 0.023×10^{-3}$

$Solubility=S= 0.023×10^{-3}×10^{3}/1 = 0.023mol/ltr$

2K

$K_2PtCl_6 ->2K^{+} +Pt^{4+}+6Cl^{-}$

$Solubility product = [2S]^{2}×[S]×[6S]^{6}$ ]

$Solubility product = 4S^{9}×6^6$

$Solubility product = 4×6^{6}×(0.023)^{9}$

$Solubility product = 3.361×10^{-10}(mol/ltr)^9$

I have taken K₂PtCl₆ instead of K₂PtCl as no such compound exist as $Pt^{-}$ does not exist.