Question

A buffer is prepared by mixing 204.0 mL of 0.452 M HCl and 0.500 L of 0.400 M sodium acetate (NaC2H3O2):

Answer 1

Answer:

pH = 4.89

0.6171g of KOH must be added

Explanation:

Sodium acetate reacts with HCl producing acetic acid as follows:

NaC2H3O2 + HCl → HC2H3O2 + NaCl

That means the moles added of HCl are moles of acetic acid produced and moles of acetate are initial moles - moles of HCl

Moles HCl = Moles acetic acid:

0.204L * (0.452mol / L) = 0.0822 moles acetic acid.

Initial moles sodium acetate:

0.500L * (0.400mol / L) = 0.200 moles

Moles sodium acetate:

0.200 moles - 0.0822 moles = 0.1172 moles sodium acetate

The pH of this buffer (Mixture of a weak acid and its conjugate base) is obtained using H-H equation:

pH = pKa + log [A-] / [HA]

Where pH is pH of the buffer,

pKa is pKa of the buffer (4.74 for acetic acid)

[A-] Moles of sodium acetate -Conjugate base-

And HA moles of acetic acid -Weak acid-

pH = 4.74 + log [0.1172 moles] [0.0822 moles]

pH = 4.89

In 0.500L the moles of the buffer are:

0.500L * (0.200moles / 0.704L) = 0.142 moles of buffer

For a pH of 0.15 units more (4.89 + 0.15 = 5.04):

5.04 = 4.74 + log [Acetate] / [Acetic acid]

0.3 = log [Acetate] / [Acetic acid]

1.9953 = [Acetate] / [Acetic acid] (1)

And as:

0.142 = [Acetate] + [Acetic acid] (2)

Replacing (2) in (1):

1.9953 = 0.142 - [Acetic acid]  / [Acetic acid]

1.9953 [Acetic acid] = 0.142 - [Acetic acid]

2.9953 [Acetic acid] = 0.142 moles

[Acetic acid] = 0.0474 moles of acetic acid must remain after the addition of KOH.

In the beginning, moles of acetic acid are:

0.0822 moles * (0.500L / 0.704L) = 0.0584 moles.

That means moles added of KOH (Equal to moles of acetic acid that react) are:

0.0584 moles - 0.0474 moles = 0.0110 moles of KOH

In grams (Molar mass KOH = 56.1g/mol):

0.0110 moles KOH * (56.1g/mol) =

0.6171g of KOH must be added