Question

50 mL of 0.1 M acetic acid is mixed with 50 mL of 0.1 M sodium acetate (the conjugate base). The Ka of acetic acid is approximately 1. 74 X 10 -5. What is the pH of the resulting solution?

Answer 1

Answer:

4.76

Explanation:

In this case, we have to start with the buffer system:

$CH_3COOH~->~CH_3COO^-~+~H^+$

We have an acid ($CH_3COOH$) and a base ($CH_3COO^-$). Therefore we can write the henderson-hasselbach reaction:

$pH~=~pKa+Log\frac{[CH_3COO^-]}{[CH_3COOH]}$

If we want to calculate the pH, we have to calculate the pKa:

$pH=-Log~Ka=4.76$

According to the problem, we have the same concentration for the acid and the base 0.1M. Therefore:

$[CH_3COO^-]=[CH_3COOH]$

If we divide:

$\frac{[CH_3COO^-]}{[CH_3COOH]}~=~1$

If we do the Log of 1:

$Log~1=~zero$

So:

$pH~=~pKa$

With this in mind, the pH is 4.76.

I hope it helps!