How to calculate the solubility of lead (II) carbonate in water, given that bicarbonate will also form?
List known reactions and equilibrium constants

\begin{align}
\ce{PbCO3 &amp;&lt;--&gt; Pb^2+ + CO3^2-} \quad \quad \quad \quad \quad K_\mathrm{sp} = 1.5\times10^{-15}\\
\ce{HCO3- &amp;&lt;--&gt; H+ + CO3^2-} \quad \quad \quad \quad \quad \quad K_\mathrm{a2} = 4.7\times10^{-11}\\
\ce{CO3^2- + H2O &amp;&lt;--&gt; HCO3- + OH-} \quad \quad \quad \quad \quad K_\mathrm{b2} = \dfrac{K_\mathrm{w}}{K_\mathrm{a2}} = 2.13 \times 10^{-4}\\
\end{align}

Start off by just evaluating quick and dirty


Assume $\ce{[Pb^{2+}] = [CO3^{2-}]}$, $\ce{[Pb^{2+}] = \sqrt{1.5\times10^{-15}} = 0.387 \times10^{-7}}$


OK, the assumption is no good and this is going to be fairly nasty...


The pH is close to the autoionization of water 
$\ce{CO3^{2-}}$ is a strong base so $\ce{HCO3^{-}}$ can't be ignored. 


GIVEN:


Starting off with pure water so $\ce{[H+] = [OH-] } = 1\times 10^{-7}$ before $\ce{PbCO3}$ is dissolved.
$\ce{[CO3^{2-}] = 0}$ and $\ce{[HCO3^-] = 0}$ before $\ce{PbCO3}$ is dissolved.


WORK

Know that


The final pH > 7 due to the formation of $\ce{HCO3^{-}}$
$\ce{[Pb^{2+}] = [CO3^{2-}] +[HCO3^-]}$
$\ce{[OH^-]_{final} - [OH^-]_{water} = [HCO3^-] }$ or $\ce{[OH^-]_{final} = 1.00\times 10^{-7} + [HCO3^-] }$


For the ratio $ \dfrac{\ce{[CO3^{2-}]}}{\ce{[HCO3^-]}}$ we can use the expression of the $K_\mathrm{a2}$ equilibrium to find that:


At pH 7 ratio = 0.00047
At pH 8 ratio = 0.0047
At pH 9 ratio = 0.047


So it would seem that $\ce{[HCO3^{2-}] &gt;&gt; [CO3^-]}$ is a reasonable assumption, which also means that $\ce{[Pb^{2+}] \approx [HCO3^-]}$.

Now we have the two equilibrium equations 

$K_\mathrm{sp} = 1.5\times10^{-15} = \ce{[Pb^{2+}][CO3^{2-}]}$

$K_\mathrm{b2} = \dfrac{K_\mathrm{w}}{K_\mathrm{a2}} = \dfrac{\ce{[HCO3^-][OH-]}}{\ce{[CO3^{2-}]}}$

which we can multiply together to get

$\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}} = \ce{[Pb^{2+}][HCO3^{-}][OH^-]}$

APPROXIMATE CUBIC SOLUTION

if we let $x = \ce{[Pb^{2+}]}$ then since $x = \ce{[Pb^{2+}] \approx [HCO3^-]}$ and $\ce{[OH-] = [HCO3^-] + 1.00\times10^{-7}}$ we have

$\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}} =(x)(x)(x + 1\times10^{-7}) =x^3 + (1\times10^{-7})x^2$

Now if we multiply both sides by $10^{21}$ we are trying to solve:

$319.149 = x^3 + x^2$

neglecting the square term we get $x=\sqrt[3]{319.149} = 6.834$, or $\ce{[Pb^{2+}] = 6.834\times10^{-7}}$.

But looking at our cubic equation again

$319.149 = x^3 + x^2 \text{ =?= } 6.834^3 + 6.834^2 = 365.821 $

which is too far off since we need values to about 1%. Solving the cubic by iteration we get $x = 6.516$


  
    Using 2 guesses then interpolation
    6.834           365.876
    6.000           252.000
    6.492           315.758
    6.517           319.257
    6.516           319.116
    Check
    $319.149 = x^3 + x^2 \text{ =?= } 6.516^3 + 6.516^2 = 319.116 $
  


which gives:

$\begin{align}
\ce{[Pb^{2+}]} &amp;= 6.516 \times10^{-7}\\
\ce{[HCO3^{-}]} &amp;= 6.516 \times10^{-7}\\
\ce{[OH^{-}]} &amp;= 7.516 \times10^{-7}\\
\end{align}$

We can calculate 

$\ce{[CO3^{2-}]} = \dfrac{K_\mathrm{sp}}{\ce{[Pb^{2+}]}} = 0.023\times10^{-7} $

$\ce{[H+]} = \dfrac{K_\mathrm{w}}{\ce{[OH-]}} = 0.133\times10^{-7}$

CHECK:

$\dfrac{\ce{[H+][CO3^{2-}]}}{\ce{[HCO3^{-}]}} = 4.695\times10^{-11}$

So to the correct number of significant figures we have

$\begin{align}
\ce{[Pb^{2+}]} &amp;= 6.5 \times10^{-7}\\
\ce{[CO3^{2-}]} &amp;= 0.023 \times10^{-7}\\
\ce{[HCO3^{-}]} &amp;= 6.5 \times10^{-7}\\
\ce{[H^{+}]} &amp;= 0.13 \times10^{-7}\text{,}\quad pH = 7.88\\
\ce{[OH^{-}]} &amp;= 7.5 \times10^{-7}
\end{align}$

EXACT FORMULIC SOLUTION 

Starting with the formula

$\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}} = \ce{[Pb^{2+}][HCO3^{-}][OH^-]}$

We let $x = \ce{[Pb^{2+}]}$ as before.

$\ce{[HCO3^-] = [Pb^{2+}] - [CO3^{2-}] =} x - \dfrac{K_\mathrm{sp}}{x} = \dfrac{x^2 - K_\mathrm{sp}}{x}$

$\ce{[OH-] = [HCO3^-]} + \sqrt{K_\mathrm{w}} = \dfrac{x^2 - K_\mathrm{sp}}{x} + \sqrt{K_\mathrm{w}} = \dfrac{x^2 +(\sqrt{K_\mathrm{w}})x - K_\mathrm{sp}}{x}$

$\begin{align}
\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}} &amp;= (x)(\dfrac{x^2 - K_\mathrm{sp}}{x})(\dfrac{x^2 +(\sqrt{K_\mathrm{w}})x - K_\mathrm{sp}}{x})\\
\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}}x &amp;=(x^2 - K_\mathrm{sp})(x^2 +(\sqrt{K_\mathrm{w}})x - K_\mathrm{sp}) \\
\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}}x &amp;= x^4 + (\sqrt{K_\mathrm{w}})x^3 - 2K_\mathrm{sp}x^2 - K_\mathrm{sp}\sqrt{K_\mathrm{w}}x + K^2_\mathrm{sp}\\
0 &amp;= x^4 + (\sqrt{K_\mathrm{w}})x^3 - 2K_\mathrm{sp}x^2 - \dfrac{K_\mathrm{w}K_\mathrm{sp}+ K_\mathrm{sp}K_\mathrm{a2}\sqrt{K_\mathrm{w}}}{K_\mathrm{a2}}x + K^2_\mathrm{sp}\\
\end{align}$

$x = 6.531198\times10^{-07}$, and other values same as Excel also...

EXACT SOLUTION WITH EXCEL

Solved first by using an excel spreadsheet (with too many significant figures, but extra digits help show that the calculations are consistent)...*

$\begin{align}
\ce{[Pb^{2+}]} &amp;= 6.531199 \times10^{-7}\\
\ce{[CO3^{2-}]} &amp;= 0.022967 \times10^{-7}\\
\ce{[HCO3^{-}]} &amp;= 6.508231 \times10^{-7}\\
\ce{[H^{+}]} &amp;= 0.133187 \times10^{-7}\text{,}\quad pH = 7.8755376\\
\ce{[OH^{-}]} &amp;= 7.508230 \times10^{-7}
\end{align}$

CHECKS:

$\ce{[H^{+}][OH-] = 0.9999986\times 10^{-14}}$

$\ce{[Pb^{2+}][CO3^{2-}]} = 1.500020\times10^{-15} $

$\dfrac{\ce{[H+][CO3^{2-}]}}{\ce{[HCO3^{-}]}} = 4.700057\times10^{-11}$