A 4.50-g quantity of a diprotic acid was dissolved in water and made up to exactly 225 mL. Calculate the molar mass of the acid if 25.0 mL of this solution required 10.5 mL of 1.00 M KOH for neutralization. Assume that both protons of the acid were titrated.

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### Problem Statement

A 4.50-g quantity of a diprotic acid was dissolved in water and made up to exactly 225 mL. Calculate the molar mass of the acid if 25.0 mL of this solution required 10.5 mL of 1.00 M KOH for neutralization. Assume that both protons of the acid were titrated.

\[ \_\_\_\_ \text{ g/mol} \]

### Explanation

In this problem, you are given a sample of diprotic acid and need to find its molar mass. A diprotic acid can release two protons per molecule. The procedure provided includes dissolving a known mass of the acid in water, titrating a portion of that solution with a standard base, and using stoichiometry to find the molar mass. 

**Key Steps to Solve the Problem:**

1. **Identify Given Values:**
   - Mass of diprotic acid = 4.50 g
   - Volume of solution prepared = 225 mL
   - Volume of solution used for titration = 25.0 mL
   - Volume of KOH used = 10.5 mL
   - Molarity of KOH = 1.00 M

2. **Understand the Neutralization Reaction:**
   The balanced chemical equation for the neutralization of a diprotic acid (H2A) with KOH is:
   \[ \text{H}_2\text{A} + 2 \text{KOH} \rightarrow \text{K}_2\text{A} + 2 \text{H}_2\text{O} \]

3. **Perform Calculations:**
   - Calculate moles of KOH used: 
     \[ \text{Moles of KOH} = \text{Volume (L)} \times \text{Molarity} \]
   - Calculate moles of the diprotic acid in the titrated sample using the stoichiometric relationship (1 mole of H2A reacts with 2 moles of KOH).
   - Calculate total moles of diprotic acid in 225 mL using the proportion given by the 25.0 mL sample.
   - Determine the molar mass by dividing the initial mass of the diprotic acid by the total moles calculated.

This problem tests your understanding
Transcribed Image Text:### Problem Statement A 4.50-g quantity of a diprotic acid was dissolved in water and made up to exactly 225 mL. Calculate the molar mass of the acid if 25.0 mL of this solution required 10.5 mL of 1.00 M KOH for neutralization. Assume that both protons of the acid were titrated. \[ \_\_\_\_ \text{ g/mol} \] ### Explanation In this problem, you are given a sample of diprotic acid and need to find its molar mass. A diprotic acid can release two protons per molecule. The procedure provided includes dissolving a known mass of the acid in water, titrating a portion of that solution with a standard base, and using stoichiometry to find the molar mass. **Key Steps to Solve the Problem:** 1. **Identify Given Values:** - Mass of diprotic acid = 4.50 g - Volume of solution prepared = 225 mL - Volume of solution used for titration = 25.0 mL - Volume of KOH used = 10.5 mL - Molarity of KOH = 1.00 M 2. **Understand the Neutralization Reaction:** The balanced chemical equation for the neutralization of a diprotic acid (H2A) with KOH is: \[ \text{H}_2\text{A} + 2 \text{KOH} \rightarrow \text{K}_2\text{A} + 2 \text{H}_2\text{O} \] 3. **Perform Calculations:** - Calculate moles of KOH used: \[ \text{Moles of KOH} = \text{Volume (L)} \times \text{Molarity} \] - Calculate moles of the diprotic acid in the titrated sample using the stoichiometric relationship (1 mole of H2A reacts with 2 moles of KOH). - Calculate total moles of diprotic acid in 225 mL using the proportion given by the 25.0 mL sample. - Determine the molar mass by dividing the initial mass of the diprotic acid by the total moles calculated. This problem tests your understanding
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