How do we find the #K_"sp"# value for #"lead nitrate"#?

1 Answer
anor277
Apr 11, 2017

You need to quote some units.........

Explanation:

We interrogate the equilibrium:

#Pb(NO_3)_2(s) rightleftharpoons Pb^(2+) + 2NO_3^-#

And #K_"sp"=[Pb^(2+)][NO_3^-]^2#

If we call the solubility of #"lead nitrate"# #=# #S#, then.......

#K_"sp"=[Pb^(2+)][NO_3^-]^2=(S)(2S)^2=4S^3#.

But you have given us NO VALUE for solubility. What are the units, #"grams per mL,"# #mol*L^-1#, #"pounds per pint"#? Of course, I could look up this value, but why should I bother if you can't be bothered.

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