Question

You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (p K a = 4.20 ) and 0.200 M sodium benzoate. How many milliliters of each solution should be mixed to prepare this buffer?

Answer 1

Answer : The volume of sodium benzoate and benzoic acid  solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.

Explanation :

Let the volume of sodium benzoate (salt) be, x

So, the volume of benzoic acid  (acid) will be, (100 - x)

Using Henderson Hesselbach equation :

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

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

$4.00=4.20+\log \left(\frac{(\frac{0.200x}{100})}{(\frac{0.100(100-x)}{100})}\right)$

x = 29.0

The volume of sodium benzoate = x = 29.0 mL

The volume of benzoic acid  (acid) = (100 - x) = (100 - 29.0) = 71 mL

Thus, the volume of sodium benzoate and benzoic acid  solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.