Answer :
To solve this problem, we need to determine how much 0.562 M KOH solution is required to completely neutralize 125.0 mL of 0.882 M H₂SO₄. Here's a step-by-step guide to find the answer:
1. Understand the Reaction:
The neutralization reaction between sulfuric acid (H₂SO₄) and potassium hydroxide (KOH) can be represented by the balanced chemical equation:
[tex]\[
\text{H}_2\text{SO}_4 + 2\text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}
\][/tex]
This equation tells us that 1 mole of H₂SO₄ reacts with 2 moles of KOH.
2. Calculate Moles of H₂SO₄:
First, we calculate the moles of H₂SO₄ in the solution. The formula for moles is:
[tex]\[
\text{moles of H}_2\text{SO}_4 = \text{Concentration (M)} \times \text{Volume (L)}
\][/tex]
Convert 125.0 mL to liters:
[tex]\[
125.0 \, \text{mL} = 0.1250 \, \text{L}
\][/tex]
Now calculate the moles of H₂SO₄:
[tex]\[
\text{moles of H}_2\text{SO}_4 = 0.882 \, \text{M} \times 0.1250 \, \text{L} = 0.11025 \, \text{moles}
\][/tex]
3. Determine Moles of KOH Required:
According to the balanced equation, 1 mole of H₂SO₄ reacts with 2 moles of KOH. Therefore, the moles of KOH required are:
[tex]\[
\text{moles of KOH} = 2 \times 0.11025 = 0.2205 \, \text{moles}
\][/tex]
4. Calculate Volume of KOH Solution:
We need to find the volume of the 0.562 M KOH solution that contains 0.2205 moles of KOH. Using the formula:
[tex]\[
\text{Volume (L)} = \frac{\text{Moles}}{\text{Concentration (M)}}
\][/tex]
[tex]\[
\text{Volume of KOH (L)} = \frac{0.2205 \, \text{moles}}{0.562 \, \text{M}} \approx 0.39235 \, \text{L}
\][/tex]
Convert this volume from liters to milliliters:
[tex]\[
0.39235 \, \text{L} \times 1000 = 392.35 \, \text{mL}
\][/tex]
We can approximate this to 392 mL.
Thus, approximately 392 mL of 0.562 M KOH solution is required to neutralize the 125.0 mL of 0.882 M H₂SO₄.
1. Understand the Reaction:
The neutralization reaction between sulfuric acid (H₂SO₄) and potassium hydroxide (KOH) can be represented by the balanced chemical equation:
[tex]\[
\text{H}_2\text{SO}_4 + 2\text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}
\][/tex]
This equation tells us that 1 mole of H₂SO₄ reacts with 2 moles of KOH.
2. Calculate Moles of H₂SO₄:
First, we calculate the moles of H₂SO₄ in the solution. The formula for moles is:
[tex]\[
\text{moles of H}_2\text{SO}_4 = \text{Concentration (M)} \times \text{Volume (L)}
\][/tex]
Convert 125.0 mL to liters:
[tex]\[
125.0 \, \text{mL} = 0.1250 \, \text{L}
\][/tex]
Now calculate the moles of H₂SO₄:
[tex]\[
\text{moles of H}_2\text{SO}_4 = 0.882 \, \text{M} \times 0.1250 \, \text{L} = 0.11025 \, \text{moles}
\][/tex]
3. Determine Moles of KOH Required:
According to the balanced equation, 1 mole of H₂SO₄ reacts with 2 moles of KOH. Therefore, the moles of KOH required are:
[tex]\[
\text{moles of KOH} = 2 \times 0.11025 = 0.2205 \, \text{moles}
\][/tex]
4. Calculate Volume of KOH Solution:
We need to find the volume of the 0.562 M KOH solution that contains 0.2205 moles of KOH. Using the formula:
[tex]\[
\text{Volume (L)} = \frac{\text{Moles}}{\text{Concentration (M)}}
\][/tex]
[tex]\[
\text{Volume of KOH (L)} = \frac{0.2205 \, \text{moles}}{0.562 \, \text{M}} \approx 0.39235 \, \text{L}
\][/tex]
Convert this volume from liters to milliliters:
[tex]\[
0.39235 \, \text{L} \times 1000 = 392.35 \, \text{mL}
\][/tex]
We can approximate this to 392 mL.
Thus, approximately 392 mL of 0.562 M KOH solution is required to neutralize the 125.0 mL of 0.882 M H₂SO₄.
