Question

You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.120 M sodium benzoate. How much of this solution should be mixed to prepare this buffer? Sum of volumes must equal 100mL. Please explain clearly and show work! Thank you :)

Answer 1

Answer – 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.