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Oksi-84 [34.3K]
3 years ago
10

Calculate the pH after the addition of 10.0 mL of 0.240 M sodium hydroxide to 50.0 mL of 0.120 M acetic acid.

Chemistry
1 answer:
Goryan [66]3 years ago
6 0

Answer:

pH = 4.58

Explanation:

The reaction of NaOH with acetic acid, CH₃COOH occurs as follows:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

<em>Moles that react:</em>

NaOH = 10mL = 0.010L * (0,240mol / L) = 0.0024 moles NaOH

CH₃COOH = 50.0mL = 0.050L * (0.120mol / L) = 0.0060 moles CH₃COOH

That means after the reaction you will have:

CH₃COOH: 0.0060 mol - 0.0024 mol = 0.0036 moles

CH₃COO⁻Na⁺: 0.0024 moles

in solution, you will have the mixture of a weak acid (Acetic acid), with its conjugate base (sodium acetate, CH₃COO⁻Na⁺). And pH of this buffer can be determined using H-H equation:

pH = pKa + log [A⁻] / [HA]

For Acetic buffer pKa = 4.76:

pH = 4.76 + log [CH₃COO⁻Na⁺] / [CH₃COOH]

<em>Where [] is molarity of each species or moles</em>

<em />

Replacing:

pH = 4.76 + log [0.0024 moles] / [0.0036 moles]

<h3>pH = 4.58</h3>

<em />

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95.6 cal
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Explanation:
Use the following equation:
q
=
m
c
Δ
T
,
where:
q
is heat energy,
m
is mass,
c
is specific heat capacity, and
Δ
T
is the change in temperature.
Δ
T
=
T
final
−
T
initial
Known
m
=
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c
Pb
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0.130
J
g
⋅
∘
C
T
initial
=
17.5
∘
C
T
final
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Δ
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∘
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Unknown
q
Solution
Plug the known values into the equation and solve.
q
=
(
125
g
)
×
(
0.130
J
g
⋅
∘
C
)
×
(
24.6
∘
C
)
=
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(rounded to three significant figures)
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1 J
=
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J
×
0.2389
cal
1
J
=
95.6 cal

(rounded to three significant figures)
95.6 cal
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