This is the exact question I faced on an exam:

Calculate the $\mathrm{pH}$ of $\pu{0.05 M}~\ce{Na2CO3}. \\(\ce{H2CO3}: K_\mathrm{a_1}= 4\times 10^{-7},~ K_\mathrm{a_2}= 4.7\times 10^{-11})$


$$\ce{Na2CO3 ->2Na+ + CO3^2-}$$

I suppose nothing that can contribute to the $\mathrm{pH}$ of a solution happens to $\ce{Na+}$ ions and we proceed with $\ce{CO3^2-}$ which has a concentration of $\pu{0.05M}$

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

To calculate $\mathrm{pH}$, I need to first figure out concentration of $\ce{OH-}$ ions, and to do so I have to know $K_\mathrm{b}$ dissociation constant for $\ce{CO3^2-}$

On equilibrium concentration of species are as follows: $$[\ce{CO3^2-}] = \pu{0.05 M}-x,[\ce{HCO3-}] = [\ce{OH-}] = x$$

So, we have:

$$\dfrac{x^2}{0.05-x}=K_\mathrm b$$

All I need to know is $K_\mathrm{b}$ value for $\ce{CO3^2-}$ ion. I tried to derive $K_\mathrm{b}$ from $K_\mathrm{a}$ values using $K_\mathrm{a} \times K_\mathrm{b} = 1\times 10^{-14}$ but apparently obtained the incorrect answer. What method should I use to do it the right way?

  • 3
    $\begingroup$ You should use that method, but you messed up somewhere. Without more work, we can't tell you what you did wrong. $\endgroup$
    – Zhe
    Commented Jul 2, 2018 at 20:59
  • $\begingroup$ @Zhe Thank you,i rechecked my work,found mistake and got the correct result. $\endgroup$ Commented Jul 2, 2018 at 21:18

1 Answer 1


The carbonate ion is the Conjugate base of the weak acid $\ce{HCO_3^-}\ (K={4.7\times10^{-11}})$, so this solution will alkaline. Given the concentration of this solution ,the pH should be sufficiently high to preclude the formation of any significant amount of $\ce{H_2CO_3}$ , so the solution of this problem as a solution of a monoprotic weak base: $\ce{CO_3^{-2} + H_2O <=> HCO_3^- + OH^-}$ $$\ce{K_b}=\frac{[OH^-][HCO_3^-]}{[CO_3^{-2}]} =\frac{K_w}{Ka}=\frac{10^{-14}}{4.7\times 10^{-11}}=\ce{10^{-3.7}}$$ Neglecting the $\ce{OH^-}$ produced by the autoprotolysis of water, it is valid to make the usual assumption that $\ce{[OH^-]}={[HCO_3^-]}$,and thus $$\dfrac{[OH^-]^2}{0.05-{[OH^-]}}=K_b= \ce{10^{-3.7}}$$ The equilibrium expression must be solved as a quadratic and yields the root $\ce{[OH^-]}$=0.00306 Which corresponds to pOH = 2.5 or pH = 11.5


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