**Titration Problem Analysis**In a titration experiment involving 13.5 mL of a 0.425 M solution of diprotic acid, H₂C₄H₄O₆ (tartaric acid), with 0.155 M LiOH, the objective is to determine the volume of base required to reach the second equivalence point. **Conceptual Explanation:**- **Diprotic Acid:** This is an acid that can donate two protons (H⁺ ions) per molecule in an aqueous solution. - **Equivalence Point:** In a titration, this is the point at which the amount of acid equals the amount of base, meaning the acid has been completely neutralized by the base.- **Second Equivalence Point:** This is reached after all protons from the diprotic acid have reacted with the base.**Calculation Strategy:**1. Calculate the moles of tartaric acid present initially.2. Since it is diprotic, multiply the moles by 2 to account for both protons being neutralized.3. Use the stoichiometry of the reaction to determine the volume of LiOH needed.By determining these values, one can calculate how the titrant (LiOH) interacts with tartaric acid to reach the desired equivalence point.

Volume of tartaric acid = 13.5 ml Molarity of tartaric acid = 0.425 M Molarity of LiOH = 0.155 M Volume of base required to reach the second equivalence point (in ml) = ? Volume of tartaric acid = 13.5 ml ...Or because 1L = 1000 ml ....so, 13.5 ml × (1L / 1000 ml ) = 0.0135 L * Molarity (M) can be written as : mol/L .Number of moles = Molarity × volume of solution in Litre ....so, Number of moles of tartaric acid : = 0.425 mol/L × 0.0135 L = 0.0057375 mol. The balance chemical reaction : H2C4H4O6(aq) + 2LiOH(aq) ---> Li2C4H4O6(aq) + 2H2O(l) 1 mol tartaric acid react with 2 mol LiOH. So number of moles of LiOH required to react with 0.0057375 mol H2C4H4O6 : = 0.0057375 mol H2C4H4O6 × (2 mol LiOH / 1 mol H2C4H4O6 ) = 0.011475 mol LiOH Volume of solution in Litre = number of moles of solute / Molarity ....so, Volume of LiOH : = 0.011475 mol / 0.155 mol/L = 0.07403 L ...Or = 0.07403 L × (1000 ml / 1L ) = 74.03 ml ..Or = 74 ml Hence, Volume of base required to reach the second equivalence point = 74 ml.

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In a titration of 13.5 mL of a 0.425 M solution of a diprotic acid H2C4H&O6 (tartaric acid) with 0.155 M LİOH, how many mL of base are required to reach the second equivalence point?

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ISBN:9781305957404

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Concept explainers

Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.

Question

**Titration Problem Analysis**

In a titration experiment involving 13.5 mL of a 0.425 M solution of diprotic acid, H₂C₄H₄O₆ (tartaric acid), with 0.155 M LiOH, the objective is to determine the volume of base required to reach the second equivalence point.

**Conceptual Explanation:**
- **Diprotic Acid:** This is an acid that can donate two protons (H⁺ ions) per molecule in an aqueous solution.
- **Equivalence Point:** In a titration, this is the point at which the amount of acid equals the amount of base, meaning the acid has been completely neutralized by the base.
- **Second Equivalence Point:** This is reached after all protons from the diprotic acid have reacted with the base.

**Calculation Strategy:**
1. Calculate the moles of tartaric acid present initially.
2. Since it is diprotic, multiply the moles by 2 to account for both protons being neutralized.
3. Use the stoichiometry of the reaction to determine the volume of LiOH needed.

By determining these values, one can calculate how the titrant (LiOH) interacts with tartaric acid to reach the desired equivalence point.

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