To determine the volume of 5% NaOH solution required to neutralize a given volume of decinormal H₂SO₄ solution, we will go through the problem step-by-step. Let's break down the process clearly to understand each part:
### Step 1: Understanding the Molarity and Volume of H₂SO₄
Decinormal H₂SO₄ refers to a 0.1 M (Molar) solution of sulfuric acid. We will denote the volume of this H₂SO₄ solution as [tex]\( V_{H₂SO₄} \)[/tex] liters.
### Step 2: Calculating the Moles of H₂SO₄
To find the moles of sulfuric acid ([tex]\( H₂SO₄ \)[/tex]), we use the relationship:
[tex]\[ \text{Moles of } H₂SO₄ = \text{Molarity of } H₂SO₄ \times \text{Volume of } H₂SO₄ \][/tex]
Given that the molarity of [tex]\( H₂SO₄ \)[/tex] is 0.1 M:
[tex]\[ \text{Moles of } H₂SO₄ = 0.1 \times V_{H₂SO₄} \][/tex]
### Step 3: Balanced Chemical Equation
The balanced chemical equation for the neutralization reaction between sulfuric acid and sodium hydroxide is:
[tex]\[ H₂SO₄ + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O} \][/tex]
From the equation, we can see that one mole of [tex]\( H₂SO₄ \)[/tex] reacts with two moles of [tex]\( NaOH \)[/tex].
### Step 4: Calculating the Moles of NaOH Required
Using the stoichiometry of the reaction, the moles of NaOH required are:
[tex]\[ \text{Moles of NaOH required} = 2 \times \text{Moles of } H₂SO₄ \][/tex]
Substitute the moles of [tex]\( H₂SO₄ \)[/tex]:
[tex]\[ \text{Moles of NaOH required} = 2 \times 0.1 \times V_{H₂SO₄} \][/tex]
Simplifying this, we get:
[tex]\[ \text{Moles of NaOH required} = 0.2 \times V_{H₂SO₄} \][/tex]
### Step 5: Converting the Weight Percentage to Molarity
5% w/v NaOH solution means there are 5 grams of NaOH in 100 mL of solution. To convert this to molarity:
- First, convert grams to moles. The molar mass of NaOH is approximately 40 g/mol.
[tex]\[
\text{Molarity of NaOH} = \frac{50 \text{ g/L}}{40 \text{ g/mol}} = 1.25 \text{ M}
\][/tex]
### Step 6: Calculating the Volume of 5% NaOH Solution
Now, we determine the volume of NaOH solution needed. We use the relationship:
[tex]\[ \text{Moles of NaOH} = \text{Molarity of NaOH} \times \text{Volume of NaOH} \][/tex]
Rearranging for the volume of NaOH:
[tex]\[ \text{Volume of NaOH} = \frac{\text{Moles of NaOH required}}{\text{Molarity of NaOH}} \][/tex]
Substitute the values:
[tex]\[ \text{Volume of NaOH} = \frac{0.2 \times V_{H₂SO₄}}{1.25} \][/tex]
Simplifying this, we get:
[tex]\[ \text{Volume of NaOH} = 0.16 \times V_{H₂SO₄} \][/tex]
### Conclusion
To neutralize [tex]\( V_{H₂SO₄} \)[/tex] liters of decinormal (0.1 M) H₂SO₄, you would require:
[tex]\[ 0.16 \times V_{H₂SO₄} \][/tex] liters of 5% NaOH solution.
Thus, the necessary volume of 5% NaOH solution is [tex]\( 0.16 \times V_{H₂SO₄} \)[/tex] liters.
