Respuesta :

ChemistryHelper2024
Ka of benzoic acid = 6.5 x 10⁻⁵ 
pKa = - log Ka = - log (6.5 x 10⁻⁵) = 4.2
Benzoic acid (C₆H₅COOH) and its conjugate base Benzoate (C₆H₅COO⁻) are considered as acidic buffer.
Using Henderson- Hasselbalch equation for calculation of pH of buffer
pH = pKa + log [tex] \frac{[C6H5COO^{-} ]}{[C6H5COOH]} [/tex]
     = 4.2  + log (0.23) / (0.115) = 4.5  

Explanation:

According to the Handerson equation,  

       pH = [tex]pK_{a} + log \frac{C_{6}H_{5}COONa}{C_{6}H_{5}COOH}[/tex]

Also, as standard value of benzoic acid's [tex]K_{a} = 6.46 \times 10^{-5}[/tex]

So,     [tex]pK_{a} = -log K_{a}[/tex]

                    = [tex]-log(6.46 \times 10^{-5})[/tex]

                    = 4.19

Therefore, pH = [tex]-log K_{a} + log \frac{C_{6}H_{5}COONa}{C_{6}H_{5}COOH}[/tex]

                   = [tex]4.19 + log \frac{0.230}{0.115}[/tex]  

                   = 4.19 + 0.30

                   = 4.49

Thus, we can conclude that the pH of given solution is 4.49.

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