If the complete neutralization of 250mL of a H3PO4 solution requires 47 mL of 2.2 M KOH, what is the molarity of the H3PO4? The balanced chemical equation is provided below. H3PO4+3KOH->K3PO4+3H₂O Do not include units in your numeric answer.
If the complete neutralization of 250mL of a H3PO4 solution requires 47 mL of 2.2 M KOH, what is the molarity of the H3PO4? The balanced chemical equation is provided below. H3PO4+3KOH->K3PO4+3H₂O Do not include units in your numeric answer.
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![**Question:**
If the complete neutralization of 250 mL of an H₃PO₄ solution requires 47 mL of 2.2 M KOH, what is the molarity of the H₃PO₄? The balanced chemical equation is provided below.
\[ \text{H}_3\text{PO}_4 + 3\text{KOH} \rightarrow \text{K}_3\text{PO}_4 + 3\text{H}_2\text{O} \]
**Instructions:**
Do not include units in your numeric answer.
**Solution Explanation:**
Given the balanced chemical reaction and the conditions, the goal is to find the molarity (M) of the phosphoric acid (H₃PO₄) solution.
1. **Balanced Equation Analysis:**
- According to the balanced chemical equation, one mole of H₃PO₄ reacts with three moles of KOH.
2. **Calculate Moles of KOH:**
- Volume (V) of KOH = 47 mL = 0.047 L
- Molarity (M) of KOH = 2.2 M
- Moles of KOH = Molarity (M) × Volume (V) = 2.2 M × 0.047 L = 0.1034 moles
3. **Stoichiometric Relationship:**
- From the balanced equation, 3 moles of KOH react with 1 mole of H₃PO₄.
- Therefore, moles of H₃PO₄ = Moles of KOH / 3 = 0.1034 moles / 3 ≈ 0.03447 moles
4. **Calculate Molarity of H₃PO₄:**
- Volume (V) of H₃PO₄ solution = 250 mL = 0.250 L
- Molarity (M) of H₃PO₄ = Moles / Volume = 0.03447 moles / 0.250 L ≈ 0.13788 M
Thus, the molarity of the H₃PO₄ solution is approximately \( 0.138 \) (rounded to three significant figures).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9de95467-8766-4670-8442-783f9816c074%2F84a54629-e8e5-4e9f-b036-8e28444e2cde%2Fhrpsaw_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
If the complete neutralization of 250 mL of an H₃PO₄ solution requires 47 mL of 2.2 M KOH, what is the molarity of the H₃PO₄? The balanced chemical equation is provided below.
\[ \text{H}_3\text{PO}_4 + 3\text{KOH} \rightarrow \text{K}_3\text{PO}_4 + 3\text{H}_2\text{O} \]
**Instructions:**
Do not include units in your numeric answer.
**Solution Explanation:**
Given the balanced chemical reaction and the conditions, the goal is to find the molarity (M) of the phosphoric acid (H₃PO₄) solution.
1. **Balanced Equation Analysis:**
- According to the balanced chemical equation, one mole of H₃PO₄ reacts with three moles of KOH.
2. **Calculate Moles of KOH:**
- Volume (V) of KOH = 47 mL = 0.047 L
- Molarity (M) of KOH = 2.2 M
- Moles of KOH = Molarity (M) × Volume (V) = 2.2 M × 0.047 L = 0.1034 moles
3. **Stoichiometric Relationship:**
- From the balanced equation, 3 moles of KOH react with 1 mole of H₃PO₄.
- Therefore, moles of H₃PO₄ = Moles of KOH / 3 = 0.1034 moles / 3 ≈ 0.03447 moles
4. **Calculate Molarity of H₃PO₄:**
- Volume (V) of H₃PO₄ solution = 250 mL = 0.250 L
- Molarity (M) of H₃PO₄ = Moles / Volume = 0.03447 moles / 0.250 L ≈ 0.13788 M
Thus, the molarity of the H₃PO₄ solution is approximately \( 0.138 \) (rounded to three significant figures).
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