Answer:
34.5 mL
Explanation:
Given, we ned to prepared buffer solution with pH = 4.00 , volume = 100.0 mL .
[C6H5COOH] = 0.100 M , [C6H5COO-] = 0.120 M , pKa = 4.20
First we need to calculate the ratio of the conjugate base and acid
We know, Henderson Hasselbalch equation
pH = pKa + log [C6H5COO-] /[C6H5COOH]
4.00 = 4.20 + log [C6H5COO-] /[C6H5COOH]
log [C6H5COO-] /[C6H5COOH] = 4.00-4.20
= - 0.20
Antilog from both side
[C6H5COO-] /[C6H5COOH] = 0.631
Now we are given the molarity of each acid and its conjugate base and we need to calculate for volume
Sum of volume = 100 mL = 0.100 L
volume of benzoic acid = x
volume of benzoate = 0.10 -x ,
so Volume of acid + volume of conjugate base = 0.100 L
[C6H5COO-] /[C6H5COOH] = 0.631
[C6H5COO-] = 0.631 * [C6H5COOH]
0.120 (0.1-x) = 0.631 *0.100x
So, x = 0.065
So, volume of benzoic acid = x = 0.0655 L
= 65.5 mL
So, volume of sodium benzoate = 0.1 -x
= 0.1-0.0655
= 0.0345 L
= 34.5 mL
So we need to mix 65.5 mL of 0.100 M benzoic acid and 34.5 mL of 0.120 M sodium benzoate form 100.0 mL of a pH=4.00 buffer solution.
