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.
Answer:
Volume benzoic acid= 77.74 mL
Volume sodium benzoate: = 22.26 mL
Explanation:
Step 1: Data given
Volume = 100.0 mL
pH = 4.00
Molarity benzoic acid = 0.100 M
pKa benzoic acid = 4.20
Molarity sodium benzoate = 0.220 M
Step 2
pH = pKa + log([base]/[acid])
4 = 4.2 + log([base]/[acid])
-0.2 = log([base]/[acid])
10^-0.2 = [base]/[acid]
0.63 = [base]/[acid]
Step 3
start with 100 mL 0.1 M benzoic acid
0.100 L* 0.1M = 0.01 moles.
Since base/acid = 0.63, add 0.63*0.01 moles base = 0.0063 moles base.
And 0.0063 moles of a 0.22 M salt solution are x mL:
0.22 moles = 1000 mL
1 mole = 1000/0.22 mL
0.0063 moles = 1000*0.0063/0.22 mL = 28.63 mL .
So a mixture of 100 mL 0.1 M bencoic acid + 28.63 mL of 0.22 benzoate has pH =4.00
Step 4: For a total volume of 100 mL we'll have:
Volume benzoic acid: 100*100/(128.63) = 77.74 mL
Volume sodium benzoate: 28.63*100/(128.63) = 22.26 mL
