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Lelu [443]
3 years ago
5

Aspirin (C9H8O4) is an acid which can be titrated with a base to determine purity. If an aspirin tablet weighing 0.615 g is titr

ated with standardized 0.1121 M KOH, the endpoint is reached after 20.52 mL of KOH have been added. What is the percentage of aspirin in the tablet
Chemistry
1 answer:
Iteru [2.4K]3 years ago
6 0

Answer:

67.4 % of C₉H₈O₄

Explanation:

To make titrations problems we know, that in the endpoint:

mmoles of acid = mmoles of base

mmoles = M . volume so:

mmoles of acid = 20.52 mL . 0.1121 M

mmoles of acid = mg of acid / PM (mg /mmoles)

Let's determine the PM of aspirin:

12.017 g/m . 9 + 1.00078 g/m . 8 + 15.9994 g/m . 4 = 180.1568 mg/mmol

mass (mg) = (20.52 mL . 0.1121 M) . 180.1568 mg/mmol

mass (mg) = 414.4 mg

We convert the mass to g → 414.4 mg . 1g / 1000mg = 0.4144 g

We determine the % → (0.4144 g / 0.615 g) . 100 = 67.4 %

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<u>Explanation:</u>

pH is the negative logarithm of hydronium ion concentration present in a solution.

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  • If the solution has low hydrogen ion concentration, then the pH will be high and the solution will be basic. The pH range of basic solution is 7.1 to 14
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To calculate the pH of the solution, we use equation:

pH=-\log[H_3O^+]     ......(1)

To calculate the pOH of the solution, we use the equation:

pH + pOH = 14                ........(2)

  • <u>For 1:</u>

We are given:

pH = 5.54

Putting values in equation 1, we get:

5.54=-\log[H_3O^+]

[H_3O^+]=2.88\times 10^{-6}M

Now, putting values in equation 2, we get:

14 = 5.54 + pOH

pOH = 8.46

The solution is acidic in nature.

  • <u>For 2:</u>

We are given:

pOH = 9.7

Putting values in equation 2, we get:

14 = 9.7 + pH

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The solution is acidic in nature.

  • <u>For 3:</u>

We are given:

pH = 7.0

Putting values in equation 1, we get:

7.0=-\log[H_3O^+]

[H_3O^+]=1.00\times 10^{-7}M

Now, putting values in equation 2, we get:

14 = 7.0 + pOH

pOH = 7.0

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  • <u>For 4:</u>

We are given:

pH = 12.9

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12.9=-\log[H_3O^+]

[H_3O^+]=1.26\times 10^{-13}M

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14 = 12.9 + pOH

pOH = 1.1

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  • <u>For 5:</u>

We are given:

pOH = 1.2

Putting values in equation 2, we get:

14 = 1.2 + pH

pH = 12.8

Now, putting values in equation 1, we get:

12.8=-\log[H_3O^+]

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The solution is basic in nature.

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We are given:

[H_3O^+]=1\times 10^{-5}M

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pH=-\log(1\times 10^{-5})

pH=5

Now, putting values in equation 2, we get:

14 = 5 + pOH

pOH = 9

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