How to calculate the pH value of a Carbonate solution?

Question

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? 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

Answer 1

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:

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

Answer 2

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 $$p\mathrm{H} = 10.97$$