So, I was doing my homework and had to write out the complete equation and net ionic equation for a combination of Lead (II) Iodate and Sodium Hydroxide.

So my equation is given as:

$\ce{Pb(IO3)2_{(aq)} + NaOH_{(aq)} -> ?}$

However my dilemma is that both of these compounds seem to be insolvable and I'm not really sure that I can either write the complete equation or even the net ionic equation, since I have no idea what to expect. If I do it on Wolfram Alpha I get the resulting products as:

$\ce{H2 + I2 + 2Na + 4O2 + Pb}$

However I have no idea if this is right/and or what's even going on here?

  • 5
    $\begingroup$ I'm not sure what Wolfram Alpha is doing; its answer is wrong! $\endgroup$
    – Ben Norris
    Oct 7, 2013 at 22:06
  • 2
    $\begingroup$ It seems to me that WolframAlpha is merely breaking the reaction down into its balanced "standard standard" compounds. Also, I'm not quite sure how sodium hydroxide is insoluble. I just dissolved a whole boatload of it less than 12 hours ago. $\endgroup$
    – chipbuster
    Oct 14, 2013 at 6:58
  • $\begingroup$ That might be a redox reaction Wolfram Alpha is giving you. $\endgroup$
    – Dissenter
    May 30, 2014 at 17:00

1 Answer 1


While lead(II) iodate is indeed sparingly soluble in water, this is not true for sodium hydroxide (111 g/100 mL at 20 °C). The reaction that occurs is a precipitation:

$$\ce{Pb(IO3)2_{(aq)} + 2NaOH_{(aq)} -> Pb(OH)2\downarrow + 2NaIO3_{(aq)}}$$

Both lead salts are not readily soluble in water; however, $\ce{Pb(OH)2}$ has a lower solubility ($K_{sp}=1.43 \times 10^{-20}$) than $\ce{Pb(IO3)2}$ with $K_{sp}=3.69 \times 10^{-13}$. Adding $\ce{NaOH}$ to an aqueous solution of lead iodate should therefore precipitate lead hydroxide, which dissolves again with excess $\ce{NaOH}$ under formation of hydroxoplumbates.


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