Question

A solution is prepared by mixing 0.12 moles of acetic acid with 0.22 moles of sodium acetate in 1.00 liters of solution. What will be the pH of the solution once equilibrium is established?

Answer 1

Answer : The pH of the solution is, 5.01

Explanation :

For acetic acid : $pK_a=4.75$

First we have to calculate the concentration of acetic acid and sodium acetate.

Concentration of acetic acid (Acid) = $\frac{Moles}{Volume}=\frac{0.12mol}{1.00L}=0.12M$

Concentration of sodium acetate (salt) = $\frac{Moles}{Volume}=\frac{0.22mol}{1.00L}=0.22M$

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

Now put all the given values in this expression, we get:

$pH=4.75+\log (\frac{0.22}{0.12})$

$pH=5.01$

Therefore, the pH of the solution is, 5.01