What is the pH of the solution when 50.0 mL of 0.20 M HI is titrated with 40.0 mL of 0.139 M Sr(OH)2? 2HI + Sr(OH)2 → Srl2 + 2H₂O pH = [?]

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**Educational Website Content**

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### Titration Problem: Determining the pH of a Solution

**Problem Statement:**
What is the pH of the solution when 50.0 mL of 0.20 M HI is titrated with 40.0 mL of 0.139 M Sr(OH)₂?

**Chemical Equation:**
\[ 2 \text{HI} + \text{Sr(OH)}_2 \rightarrow \text{SrI}_2 + 2 \text{H}_2\text{O} \]

**Solution Calculation:**
\[ \text{pH} = \boxed{\ ?\ } \]

**Hint:**
Be careful of the mole ratios!

**Interactive Component:**
Below is an input field for you to provide the pH after considering the reaction and calculations:
\[ \text{pH at 40.0 mL base:} \underline{\hspace{150px}} \ \text{Enter} \]

**Detailed Explanation of Diagram:**
The diagram features a balanced chemical equation, showing that two moles of Hydrogen Iodide (HI) react with one mole of Strontium Hydroxide \(\text{Sr(OH)}_2\) to produce one mole of Strontium Iodide (\(\text{SrI}_2\)) and two moles of water (\(\text{H}_2\text{O}\)).

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**Steps to Calculate the pH:**

1. **Calculate moles of HI and Sr(OH)₂:**
   - Moles of HI: \( M \times V = 0.20 \, \text{M} \times 0.050 \, \text{L} = 0.010 \, \text{moles} \)
   - Moles of Sr(OH)₂: \( M \times V = 0.139 \, \text{M} \times 0.040 \, \text{L} = 0.00556 \, \text{moles} \)

2. **Determine the limiting reagent:**
   - 2 moles of HI react with 1 mole of Sr(OH)₂
   - We need twice the moles of HI compared to Sr(OH)₂, i.e., 0.01112 moles of HI are required to react with 0
Transcribed Image Text:**Educational Website Content** --- ### Titration Problem: Determining the pH of a Solution **Problem Statement:** What is the pH of the solution when 50.0 mL of 0.20 M HI is titrated with 40.0 mL of 0.139 M Sr(OH)₂? **Chemical Equation:** \[ 2 \text{HI} + \text{Sr(OH)}_2 \rightarrow \text{SrI}_2 + 2 \text{H}_2\text{O} \] **Solution Calculation:** \[ \text{pH} = \boxed{\ ?\ } \] **Hint:** Be careful of the mole ratios! **Interactive Component:** Below is an input field for you to provide the pH after considering the reaction and calculations: \[ \text{pH at 40.0 mL base:} \underline{\hspace{150px}} \ \text{Enter} \] **Detailed Explanation of Diagram:** The diagram features a balanced chemical equation, showing that two moles of Hydrogen Iodide (HI) react with one mole of Strontium Hydroxide \(\text{Sr(OH)}_2\) to produce one mole of Strontium Iodide (\(\text{SrI}_2\)) and two moles of water (\(\text{H}_2\text{O}\)). --- **Steps to Calculate the pH:** 1. **Calculate moles of HI and Sr(OH)₂:** - Moles of HI: \( M \times V = 0.20 \, \text{M} \times 0.050 \, \text{L} = 0.010 \, \text{moles} \) - Moles of Sr(OH)₂: \( M \times V = 0.139 \, \text{M} \times 0.040 \, \text{L} = 0.00556 \, \text{moles} \) 2. **Determine the limiting reagent:** - 2 moles of HI react with 1 mole of Sr(OH)₂ - We need twice the moles of HI compared to Sr(OH)₂, i.e., 0.01112 moles of HI are required to react with 0
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