# How to calculate the solubility of lead (II) carbonate in water, given that bicarbonate will also form?

Calculate the solubility of lead (II) carbonate, $\ce{PbCO3}$, ($K_{\mathrm{sp}} = 1.5 \times 10^{-15}$) in water, given that carbonate is a stronger base than water, so will also form bicarbonate. When bicarbonate acts as an acid: \begin{align} \ce{HCO3- &<--> H+ + CO3^2-} & K_{\mathrm{a}} &= 4.7 \times 10^{-11}. \end{align}

I think the equation is, in light of airhuff's suggestion, $$\ce{PbCO3 -> Pb^2+ + CO3^2-}.$$ What I'm most confused about is how the bicarbonate and $K_\mathrm{a}$ comes into play in this question. It makes a traditional $K_\mathrm{sp}$ question a little trickier.

Possible answer?: \begin{align} \ce{PbCO3 &<--> Pb^2+ + CO3^2-} &\implies K_1 &= K_\mathrm{sp} = 1.5\times10^{-15}\\ \ce{CO3^2- + H2O &<--> HCO3- + OH-} &\implies K_2\\ \ce{PbCO3 + H2O &<--> Pb^2+ + HCO3- + OH-} &\implies K_3\\ \ce{HCO3- &<--> H+ + CO3^2-} &\implies K_4 &= K_\mathrm{a} = 4.7\times10^{-11}\\[3ex] K_3 &= K_1 \cdot K_2\\ K_3 &= \frac{K_\mathrm{w}}{K_4}\\ K_2 &= \frac{1\times10^{-14}}{4.7\times10^{-11}}\\ &= 2.1\times10^{-9}\\ K_3 &= (1.5\times10^{-15})\cdot(2.1\times10^{-4})\\ &= 3.15\times10^{-19}\\ S_{\ce{PbCO3}} &= [\ce{Pb^2+}][\ce{HCO3-}]\\ &=[\ce{OH-}]\\ K_3 &= [\ce{Pb^2+}][\ce{HCO3-}][\ce{OH-}]\\ &= S^3\\ S &= \sqrt[3]{K_3} (3.15\times10^{-19})\\ &= 6.8\times10^{-7}~\mathrm{M}\\ \end{align}

Does the way I solved it look okay?

• I'm pretty sure this will be closed shortly if you don't show some more work. How about writing out the chemical reaction for the solubility of $\ce{PbCO3}$? Take a shot at combining the reactions an calculating the solubility? I've seen you post some stuff before and I think you can at least make a good effort toward solving this one. Mar 2, 2017 at 18:01
• I've tried to solve this question and reached a final answer to what I think may be right?
– Greg
Mar 6, 2017 at 0:23
• @orthocresol , I think I got this pretty complex formatting right, other than that I couldn't figure out how to do a cubed root so did a power to the 1/3. You've been helpful in getting all this through my head and I just wanted to bounce this one off you...thx for any input. This message will self destruct rather than be left here ;) Mar 6, 2017 at 1:12
– Greg
Mar 6, 2017 at 1:15
• Greg, I just nominated this question to be reopened based on all the work you've put into it. This doesn't mean it will be reopened, just that I think it should be. Regarding the formatting, it's much easier to read when properly formatted. One of the reasons I did all that work to re-format was to increase the likelihood that it would be re-opened. I don't have the time to look closer at it now, but it looks like you're on the right track anyway. Mar 6, 2017 at 2:29

List known reactions and equilibrium constants

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 $\ce{[HCO3^{2-}] >> [CO3^-]}$ 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+}]} &= 6.516 \times10^{-7}\\ \ce{[HCO3^{-}]} &= 6.516 \times10^{-7}\\ \ce{[OH^{-}]} &= 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+}]} &= 6.5 \times10^{-7}\\ \ce{[CO3^{2-}]} &= 0.023 \times10^{-7}\\ \ce{[HCO3^{-}]} &= 6.5 \times10^{-7}\\ \ce{[H^{+}]} &= 0.13 \times10^{-7}\text{,}\quad pH = 7.88\\ \ce{[OH^{-}]} &= 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}} &= (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 &=(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 &= 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 &= 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+}]} &= 6.531199 \times10^{-7}\\ \ce{[CO3^{2-}]} &= 0.022967 \times10^{-7}\\ \ce{[HCO3^{-}]} &= 6.508231 \times10^{-7}\\ \ce{[H^{+}]} &= 0.133187 \times10^{-7}\text{,}\quad pH = 7.8755376\\ \ce{[OH^{-}]} &= 7.508230 \times10^{-7} \end{align}

CHECKS:

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

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

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