This can be done by dissolving PtCl4 into HCl, but if I want 5mM of H2PtCl6, then what amounts of PtCl4 and HCl should I use?
If the reaction is as follows:
PtCl4 + 2HCl = H2PtCl6
Then for 5mM H2PtCl6 you need to mix 5 mM of PtCl4 and 10mM of HCl.
Both of comments are correct, I would be one remark only, 10 mM HCl will not probably be enough to dissolve PtCl4 without residue due to hydrolysis. If you want 5 mM H2PtCl6, you have to dissolve really 5 mM PtCl4 in HCl soln. (more than 10 mM) enough to dissolve it completely. Excess of HCl needs to stabilize the H2PtCl6 because of acidic nature (hydrolysis) of H2PtCl6.
thank you all of you.
ok tell me sir it is more relevent to use commercially available H2PtCl6?
OK prof V.S Muralidharan sir....but in aldrich H2PtCl6 ,the water molecules are there..what type of the product I should use for my counter electrode.
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