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atroni [7]
3 years ago
5

A 25.0 ml sample of a 0.100 m solution of acetic acid is titrated with a 0.125 m solution of naoh. Calculate the ph of the titra

tion mixture after 10.0
Chemistry
1 answer:
Sedbober [7]3 years ago
5 0

Answer:

1) after 10.0 mL base added  the pH is 4.75

2) after 20.0 mL base added  the pH is  8.74

3)  after 30.0 mL base added  the pH is 12.36

Explanation:

CH₃COOH + NaoH --------------> CH₃COONa + H₂O

1) after 10.0 mL base added

millimoles of NaOH = 10.0 x 0.125 = 1.25

2.50 - 1.25 = 1.25 means 50% titration complete.

at half equivalence point pH = pKa

pH = 4.75

2) millimoles of NaOH added = 20.0 x 0.125 = 2.50

equivalence point

[CH₃COONa] = 2.50 / 20 + 25 = 0.055 M

pH = 1/2 [pKw + pKa + log C]

pH = 1/2 [14 + 4.75 + log 0.055]

pH = 8.74

3) millimoles of NaOH added = 30.0 x 0.125 = 3.75

3.75 - 2.50 = 1.25 millimoles NaOH left

[NaOH] = 1.25 / 30 + 25 = 0.023 M

as NaOH is strong base

[OH-] = [NaOH] = 0.023 M

pOH = - log [OH-]

pOH = - log [0.023]

pOH = 1.64

pH = 14 - 1.64

pH = 12.36

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                     1. 10.66 moles of Al

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                    The balance chemical equation is as follow,

                                      4 Al + 3 O₂ → 2 Al₂O₃

<h3>1.</h3>

According to balance chemical equation,

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According to balance chemical equation,

                       4 moles of Al produced  =  2 moles of Al₂O₃

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                     5 moles of Al will produce  =  X moles of Al₂O₃

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The balance chemical equation for the oxidation of carbon disulfide is as follow;

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Step 1: <u>Calculate Moles of CS₂ as;</u>

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                  Moles  =  1.12 moles of CS₂

Step 2: <u>Find out moles of SO₂ as;</u>

According to balance chemical equation,

                       1 mole of CS₂ produced  =  2 moles of SO₂

So,

                     1.12 moles of CS₂ will produce  =  X moles of SO₂

Solving for X,

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                  Mass  =  2.24 mol × 64.06 g/mol

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