Question

You have 0.500 L of an 0.250 M acetate buffer solution (i.e. [HC₂H₃O₂] + [C₂H₃O₂⁻] = 0.250 M) at pH 3.50. How many mL of 1.000 M NaOH must you add in order to change the pH to 5.07? Acetic acid has a pKa of 4.74.

Answer 1

80mL of 1.00M NaOH

Step-by-step explanation:

Using H-H equation, we can determine oH of a buffer as acetate buffer. First, we need to determine amount of acetate ion and acetic acid at pH 3.50 and 5.07. Then, with the reaction of NaOH with acetic acid we can find the amount of 1.00M NaOH that must be added:

At pH 3.50:

pH = pka + log [C₂H₃O₂⁻] / [HC₂H₃O₂]

3.50 = 4.74 + log [C₂H₃O₂⁻] / [HC₂H₃O₂]

0.057544 = [C₂H₃O₂⁻] / [HC₂H₃O₂] (1)

Using and replacing in (1):

[HC₂H₃O₂] + [C₂H₃O₂⁻] = 0.250 M

[HC₂H₃O₂] + 0.057544[HC₂H₃O₂] = 0.250 M

1.057544 [HC₂H₃O₂] = 0.250M

[HC₂H₃O₂] = 0.2364M * 0.500L = 0.1182 moles of acetic acid at first pH

At pH 5.07:

pH = pka + log [C₂H₃O₂⁻] / [HC₂H₃O₂]

5.07 = 4.74 + log [C₂H₃O₂⁻] / [HC₂H₃O₂]

2.13796= [C₂H₃O₂⁻] / [HC₂H₃O₂] (1)

Using and replacing in (1):

[HC₂H₃O₂] + 2.13796[HC₂H₃O₂] = 0.250 M

3.13796 [HC₂H₃O₂] = 0.250M

[HC₂H₃O₂] = 0.07967M * 0.500L = 0.0398 moles of acetic acid at first pH

Now, NaOH reacts with HC₂H₃O₂ as follows:

NaOH + HC₂H₃O₂ → NaC₂H₃O₂ + H₂O

As moles of acetic acid decreases from 0,1198 moles - 0,0398 moles = 0,08 moles of acetic acid are consumed = 0,08 moles of NaOH

0,08 mol NaOH * (1L / 1mol) = 0,08L of 1.00M NaOH =

80mL of 1.00M NaOH