# Calculating the pH of a solution of sodium carbonate

I've seen similar questions on this site, but I'm having trouble as my book said something else. For a $\pu{0.09989 M}$ solution of $\ce{Na2CO3}$, my book said:

$$\ce{Na2CO3 + 2H2O -> H2CO3 + 2Na+ + 2OH-}$$

So , $\mathrm{pH} = 14 + \log(2 \times 0.09989) = 13.3006$ .

The problem is that I haven't been given a value of $K_\mathrm b$ as I saw in other questions on the site. We usually don't use any of them if they're not supplied in the question. What should I do?

• You should apply the correct formula of hydrolysis of salts( here the salt is of strong base and weak acid) which is, $$\ce{pH = 7 + \frac{pK_a(HCO_3^-)}{2} + \frac{log (c)}{2}}$$ You should look for the correctness, justify by your own which should be correct. You should be confident enough to say the thing wrong if it seems wrong by all possible logics even though it is written in book. By this formula you will find that the answer is close to $11.66$. – Soumik Das Apr 24 '18 at 18:04
• If you are given, say $K_b$, then you need to know that $K_b$ and $K_a$ are related as $K_w=K_aK_b$ . – porphyrin Apr 24 '18 at 21:29
• chemistry.stackexchange.com/questions/99012/… – Adnan AL-Amleh Aug 27 '18 at 16:45