Calculating pH of 0.1 M NaHCO3 with Ka values | Salt Hydrolysis Formula

[atavistic]

Homework Statement

Find pH of $$0.1M NaHCO_3$$

$$Ka_1 (H_2CO_3) = 4.3 * 10^{ - 7} , Ka_2 (H_2CO_3) = 5.61 * 10^{ - 11}$$

Homework Equations

$$pH = \frac {1}{2}(14 + pKa_1 + logC)$$

The Attempt at a Solution

Now $$K_h = \frac {Kw}{Ka_1} = 2.3 * 10^{ - 8} > > Ka2$$

So I neglect dissociation and use salt hydrolysis formula.

But this gives wrong answer.

 

[Borek]

pH of amphiprotic salt.

 

[Blueshift5]

I would first double check the calculations and equations being used to ensure they are correct. If the answer is still wrong, I would consider other factors that could affect the pH, such as temperature, ionic strength, and any other ions present in the solution. I would also consider the pH range being calculated and if it falls within the range where the salt hydrolysis formula is applicable. If all factors have been taken into account and the answer still seems incorrect, I would consult with colleagues or a trusted source to confirm the result.