Chemistry

Write the balanced chemical equation for each of these reactions. Include phases. When aqueous sodium hydroxide is added to a solution containing lead(II) nitrate, a solid precipitate forms. However, when additional aqueous hydroxide is added the precipitate redissolves forming a soluble [Pb(OH)4]^2–(aq) complex ion.

I got the first one, but I don't get the second one.

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  1. 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].

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  2. 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.

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  3. 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.

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  4. 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.

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  5. The answer they want is
    Pb(OH)2(S) + 2OH-(aq)==> [Pb(OH)4]2-(aq)

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  6. should be correct

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