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

This is the exact question I faced on an exam. Calculate the pH of $$\pu{0.05 M}\ \ce{Na2CO3}\ (\ce{H2CO3}: K_\mathrm{a,1}= 4\times 10^{-7},\ K_\mathrm{a,2}= 4.7\times 10^{-11})$$

### Solution

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

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

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

To calculate 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?

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

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