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]
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) =
