Posted by Sarah on Friday, May 11, 2012 at 8:21pm.
Pb^2+(aq) + 2OH^-(aq) ==> Pb(OH)2(s)
Pb(OH)2(s) + 2OH^-(aq) ==> Pb(OH)4^2-(aq)
Note: Have you encountered the word amphoteric yet? Pb(OH)2 is amphoteric; it can act as a base (it will dissolve in HNO3 to form a salt + H2O) or an acid (it reacts with a base) to form a salt [Na2Pb(OH)4 + H2O].
For the first one I put
2NaOH (aq) + Pb(NO3)2 (aq) ==> Pb(OH)2 (s) + 2NaNO3 (aq)
and for the second
Pb(OH)2 (s) + 2NaOH (aq) ==> Na2Pb(OH)4 (aq)
and it keeps saying the second one is wrong.
I don't understand what's incorrect.
These amphoteric salts get cluttered up with different ways of writing them which is why I prefer the ionic equation which i wrote for you; i.e., Pb(OH)4^2-(aq). This may be one such case. At least it's worth a try.
Try writing it this way.
Pb(OH)2(s) + 2NaOH(aq) ==> Na2PbO2(aq) + 2H2O(l)
That's the same equation but rearranged slightly.
Let me know if the Na2PbO2 works. I try to keep up with these data bases. Frankly, I think the program should be programmed to accept either answer. I accept both in my classes.
The answer they want is
Pb(OH)2(S) + 2OH-(aq)==> [Pb(OH)4]2-(aq)
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