Answer:
A solution is prepared by mixing 250 mL of 1.00 M
CH3COOH with 500 mL of 1.00 M NaCH3COO.
What is the pH of this solution?
(Ka for CH3COOH = 1.8 × 10−5 )
Explanation:
This is a case of a neutralization reaction that takes place between acetic acid, CH
3
COOH
, a weak acid, and sodium hydroxide, NaOH
, a strong base.
The resulting solution pH, depends if the neutralization is complete or not. If not, that is, if the acid is not completely neutralized, a buffer solution containing acetic acid will be gotten, and its conjugate base, the acetate anion.
It's important to note that at complete neutralization, the pH of the solution will not equal 7
. Even if the weak acid is neutralized completely, the solution will be left with its conjugate base, this is the reason why the expectations of its pH is to be over 7
.
So, the balanced chemical equation for this reaction is the ionic equation:
CH
3
COOH
(aq] + OH
−
(aq] → CH
3
COO
−
(aq] + H
2
O
(l]
Notice that:
1 mole of acetic acid will react with: 1 mole of sodium hydroxide, shown here as hydroxide anions, OH
−
, to produce 1 mole of acetate anions:
CH
3
COO
−
To determine how many moles of each you're adding
, the molarities and volumes of the two solutions are used:
c = n / V ⇒ n = c ⋅ V
n acetic = 0.20 M ⋅ 25.00 ⋅ 10 −
3
L = 0.0050 moles CH3
COOH
and
n hydroxide = 0.10 M ⋅ 40.00 ⋅ 10
−
3
L = 0.0040 moles OH
−
There are fewer moles of hydroxide anions, so the added base will be completely consumed by the reaction.
As a result, the number of moles of acetic acid that remain in solution is:
n acetic remaining = 0.0050 − 0.0040 = 0.0010 moles
The reaction will also produce 0.0040 moles of acetate anions.
This is, then a buffer and the Henderson-Hasselbalch equation is applied to find its pH
:
pH = p
K
a + log (
[
conjugate base
] / [
weak acid
]
)
Use the total volume of the solution to find the new concentrations of the acid and of its conjugate base
.
V
total = V
acetic + V
hydroxide
V
total = 25.00 mL + 40.00 mL = 65.00 mL
Thus the concentrations will be
:
[
CH
3
COOH
] = 0.0010 moles / 65.00 ⋅ 10
−
3
L = 0.015385 M
and
[
CH
3
COO
−
] = 0.0040 moles / 65 ⋅ 10
−
3
L = 0.061538 M
The p
K
a of acetic acid is equal to 4.75
Thus the pH of the solution will be:
pH = 4.75 + log
(
0.061538
M / 0.015385
M
)
pH = 5.35
