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Chapter 1: Problem 5

How would you prepare 3 liters of \(6 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution from the concentrated acid available? (Sp. Gr. = \(1.84,100 \%\) purity Mol. Weight \(\mathrm{H}_{2} \mathrm{SO}_{4}=98\) ) a. take \(239 \mathrm{~mL}\) acid and dilute to 3 liters b. take \(441 \mathrm{~mL}\) acid and dilute to 3 liters c. take \(479 \mathrm{~mL}\) acid and dilute to 3 liters d. take \(882 \mathrm{~mL}\) acid and dilute to 3 liters e. take \(958 \mathrm{~mL}\) acid and dilute to 3 liters

Short Answer

Expert verified
Option (c) is correct: take 479 ml of acid and dilute to 3 liters.

Step by step solution

01

Determine the Equivalent Weight

Calculate the equivalent weight of \(\text{H}_2\text{SO}_4\). The equivalent weight is found by dividing the molar mass by the number of hydrogen ions (H+ ions) that the acid can donate. For \(\text{H}_2\text{SO}_4\), it can donate two protons, so divide the molar mass by 2: \[\text{Equivalent Weight} = \frac{98}{2} = 49 \]
02

Calculate the Concentration of the Desired Solution

Next, calculate the normality of the concentrated solution. Given that the density (specific gravity) is 1.84 and the acid is 100% pure, use the formula: \[ \text{Normality (N)} = \frac{\text{specific gravity} \times 1000}{\text{equivalent weight}} = \frac{1.84 \times 1000}{49} = 37.55 \]
03

Apply Dilution Formula

To determine the volume of concentrated acid needed, use the dilution formula: \(\text{N}_1 \text{V}_1 = \text{N}_2 \text{V}_2\). Here, \(\text{N}_1\) is 37.55 (concentration of concentrated acid), \(\text{V}_1\) is the volume of concentrated acid needed, \(\text{N}_2\) is 6 (desired normality), and \(\text{V}_2\) is 3 liters: \[ 37.55 \times \text{V}_1 = 6 \times 3 \]
04

Solve for \(\text{V}_1\)

Rearrange the equation to solve for \(\text{V}_1\): \[ \text{V}_1 = \frac{6 \times 3}{37.55} = 0.479 \text{ liters} = 479 \text{ ml} \]
05

Verify the Correct Answer

From the given options, the correct volume (479 ml) matches option (c). Therefore, take 479 ml of concentrated \(\text{H}_2\text{SO}_4\) and dilute it to make up a total volume of 3 liters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equivalent Weight Calculation
Understanding equivalent weight is crucial when preparing chemical solutions. For sulfuric acid (\text{H}_2\text{SO}_4\text{), the equivalent weight is determined by dividing the molar mass by the number of hydrogen ions (H+ ions) it can donate. Each molecule of \text{H}_2\text{SO}_4\text{ can donate 2 hydrogen ions. Sulfuric acid has a molar mass of 98 g/mol, meaning its equivalent weight is \( \frac{98}{2} = 49 \text{g/equivalent}\). This value will be necessary for various calculations, including normality and dilution.
Normality
Normality (N) is a measure of concentration equivalent to the number of gram equivalents of solute per liter of solution. For sulfuric acid, knowing the specific gravity and purity helps calculate its normality. Given sulfuric acid with a specific gravity of 1.84 and 100% purity, the normality can be found using: \[ \text{Normality} = \frac{\text{specific gravity} \times 1000}{\text{equivalent weight}} \ = \frac{1.84 \times 1000}{49} = 37.55N\]. This indicates a very concentrated acid.
Dilution Formula
To achieve a desired normality from a more concentrated solution, the dilution formula \( \text{N}_1 \text{V}_1 = \text{N}_2 \text{V}_2 \) can be used. Here, \( \text{N}_1 \) (initial normality) is 37.55, \( \text{V}_1 \) (volume of concentrated solution needed) is unknown, \( \text{N}_2 \) (desired normality) is 6, and \( \text{V}_2 \) (final volume) is 3 liters. Plugging in these values, we get: \[ 37.55 \times \text{V}_1 = 6 \times 3 \text{L} \rightarrow \text{V}_1 = \frac{6 \times 3}{37.55} = 0.479 \text{ L} = 479 \text{ mL}\]. Thus, 479 mL of the concentrated sulfuric acid should be diluted to achieve 3 liters of a 6N solution.

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