How to calculate the pH value of a Carbonate solution?

I read the similar questions suggested when submitting my question but they didn't help me [How can one calculate the pH of a solution?](https://chemistry.stackexchange.com/questions/4010/how-can-one-calculate-the-ph-of-a-solution) [How to calculate pH of the Na2CO3 solution given ambiguous Ka values](https://chemistry.stackexchange.com/questions/99012/how-to-calculate-ph-of-the-na2co3-solution-given-ambiguous-ka-values)




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We are asked to calculate the pH value of a Carbonate Solution with a concentration of 0.005 mol per liter


They provide the numerical solution as 10.97, but not how to get it




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Carbonate has the molecular form CO3^(2-)


I know how to calculate the pH of a strong base (which Carbonate seems to be)


pH + pOH = 14 so pH = 14 - pOH, and pOH = -log\_10 (concentration = 0.005 mol per liter)


But I just don't get the same result as they give


I even assumed Carbonate is a weak base, and so used pKb = pOH^2/0.005 with pKb = 3.60 for Carbonate which I found online, and then solved for pOH and used pH = 14 - pOH, but even then I don't get their solution


I know that in water, Carbonate becomes Bicarbonate ( CO3^(2-) + H2O --> HCO3- + OH-), which then becomes Carbonic Acid (HCO3- + H2O --> H2CO3 + OH-)


All other exercises about pH they gave us, I always found the same result as they do. But for Carbonate, I just don't see what am I doing wrong. What am I missing ?


Thank you so much for your help




The equilibrium system of interest is:


$$\ce{CO3^{2-}(aq) +H2O(l)<=>HCO3^{-}(aq) +OH-(aq)}$$


Let:


A represent $\ce{CO3^{2-}}$


B represent $\ce{HCO3^{-}}$


C represent $\ce{OH^{-}}$


The $pK\_b$ of $\ce{CO3^{2-}}$ at 25°C is approximately 3.67.


Initially, only $\ce{CO3^{2-}}$ and water are present. As an approximation, water is not included in the equilibrium constant expression, so at equilibrium we have:


$$C\_A=C\_{Ao}-x=0.005-x$$


$$C\_B=C\_{Bo}+x=x$$


$$C\_C=C\_{Co}+x=x$$


Which can be substituted into the equilibrium constant:


$$K\_b=\frac{x^2}{0.005-x}=10^{-3.67}$$


Solving for $x$:


$$C\_C=x=9.33\times 10^{-4}$$


The pOH of this solution would be:


$$\pu{pOH}=\pu{-log}\;C\_C=\pu{-log}\;(9.33\times 10^{-4})=3.03$$


Finally, the pH of the solution is:


$$\pu{pH}=14-\pu{pOH}=14-3.03=10.97$$


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The pH of a diluted solution of a weak base, like $\ce{CO3^{2-}}$ is given in the tables by the following formula $${p\mathrm{H} = \frac{1}{2} ( 14 + p\mathrm{K}\_a + \mathrm{log} c\_b)}$$ The acid $\ce{HCO3^{-}}$ has a $p\mathrm{K}\_a$ value equal to $10.25$. Substituting this numerical value, plus $c\_b = 0.004$M into preceding formula, gives $$p\mathrm{H} = 10.97$$


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