# How to calculate pH of the Na2Co3 solution given ambiguous Ka values

This is the exact question i faced on 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 solution happens to $\ce{Na+}$ ions and we proceed with $\ce{CO3^2-}$ which has 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 $K_b$ 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:

$$\dfrac{x^2}{0.05-x}=K_b$$

All I need to know is Kb value for $\ce{CO3^2-}$ ion. I tried to derive $K_b$ 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 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