A 35.0 mL solution of Sr(OH)2 is neutralized with 29.5 mL of 0.200 M HNO, . What is the concentration of the original Sr(OH), solution? 2

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**Problem: Neutralization of Sr(OH)₂ with HNO₃**

**Question 7 of 12**

A 35.0 mL solution of Sr(OH)₂ is neutralized with 29.5 mL of 0.200 M HNO₃. What is the concentration of the original Sr(OH)₂ solution?

**Solution Details:**

To solve this problem, use the concept of neutralization where the number of moles of acid equals the number of moles of base at the equivalence point. The balanced chemical equation for the reaction is:

\[ \text{Sr(OH)}_2 + 2\text{HNO}_3 \rightarrow \text{Sr(NO}_3)_2 + 2\text{H}_2\text{O} \]

From the equation, 1 mole of Sr(OH)₂ reacts with 2 moles of HNO₃. 

1. Calculate moles of HNO₃:
   - Volume = 29.5 mL = 0.0295 L
   - Molarity = 0.200 M
   - Moles = Molarity × Volume = 0.200 × 0.0295 = 0.00590 moles

2. Calculate moles of Sr(OH)₂:
   - From the equation, moles of Sr(OH)₂ = 0.00590 / 2 = 0.00295 moles

3. Calculate concentration of Sr(OH)₂:
   - Volume = 35.0 mL = 0.0350 L
   - Concentration = Moles / Volume = 0.00295 / 0.0350 = 0.0843 M

Thus, the concentration of the original Sr(OH)₂ solution is 0.0843 M.

**Note:** For educational purposes, carefully analyze the stoichiometry of the balanced equation to understand the mole relationships.
Transcribed Image Text:**Problem: Neutralization of Sr(OH)₂ with HNO₃** **Question 7 of 12** A 35.0 mL solution of Sr(OH)₂ is neutralized with 29.5 mL of 0.200 M HNO₃. What is the concentration of the original Sr(OH)₂ solution? **Solution Details:** To solve this problem, use the concept of neutralization where the number of moles of acid equals the number of moles of base at the equivalence point. The balanced chemical equation for the reaction is: \[ \text{Sr(OH)}_2 + 2\text{HNO}_3 \rightarrow \text{Sr(NO}_3)_2 + 2\text{H}_2\text{O} \] From the equation, 1 mole of Sr(OH)₂ reacts with 2 moles of HNO₃. 1. Calculate moles of HNO₃: - Volume = 29.5 mL = 0.0295 L - Molarity = 0.200 M - Moles = Molarity × Volume = 0.200 × 0.0295 = 0.00590 moles 2. Calculate moles of Sr(OH)₂: - From the equation, moles of Sr(OH)₂ = 0.00590 / 2 = 0.00295 moles 3. Calculate concentration of Sr(OH)₂: - Volume = 35.0 mL = 0.0350 L - Concentration = Moles / Volume = 0.00295 / 0.0350 = 0.0843 M Thus, the concentration of the original Sr(OH)₂ solution is 0.0843 M. **Note:** For educational purposes, carefully analyze the stoichiometry of the balanced equation to understand the mole relationships.
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