# 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? How to calculate pH of the Na2CO3 solution given ambiguous Ka values

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

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:

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

• Thanks a lot for your help ! It makes a lot more sense now. The methodology you used is for a weak base, I must have done something wrong somewhere, maybe I forgot to raise 10 to the -3.67 power to get Kb Nov 26, 2022 at 1:01

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