Join the conversation on IDNLearn.com and get the answers you seek from experts. Our Q&A platform offers reliable and thorough answers to help you make informed decisions quickly and easily.
Sagot :
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.
### 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.
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. Find the answers you need at IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.
