Question

What is the pH of a buffer in which the concentration of benzoic acid, C6H5COOH, is 0.040 M and the concentration of sodium benzoate, NaC6H5COO, is 0.015 M ? Ka of C6H5COOH is 6.30 x 10-5

Answer 1

Answer:

pH=4.63

Explanation:

The firt step is identify the Acid and the conjugated base:

HA <=> $A^-$ +  $H^+$

So, for this case we will have:

$C_6H_5COOH$ <=>  $C_6H_5COO^-$ +  $H^+$

Then:

HA = $C_6H_5COOH$

$A^-$=$C_6H_5COO^-$

With this in mind we can use the henderson-hasselbach equation:

pH = pKa + Log $\frac{A^-}{HA}$

We can calculate the pKa first

$pKa~=~-~Log~(6.30x10^-^5)~=~4.2$

Then we can put the values in the equation:

pH = 4.2 + Log($\frac{0.015}{0.04}$)

pH= 4.63