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Citric acid, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7},\) is a triprotic acid. It occurs naturally in citrus fruits like lemons and has applications in food flavouring and preservatives. A solution containing an unknown concentration of the acid is titrated with KOH. It requires \(23.20 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{KOH}\) solution to titrate all three acidic protons in \(100.00 \mathrm{~mL}\) of the citric acid solution. Write a balanced net ionic equation for the neutralization reaction, and calculate the molarity of the citric acid solution.

Short Answer

Expert verified
\( 0.03867 \mathrm{M} \), Net ionic: \( \mathrm{C}_6 \mathrm{H}_5 \mathrm{O}_7^{3-} + 3 \mathrm{OH}^- \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{O}_7^{3-} + 3 \mathrm{H}_2 \mathrm{O} \)

Step by step solution

01

Analyze the Acid-Base Reaction

Citric acid, \( \mathrm{C}_6 \mathrm{H}_8 \mathrm{O}_7 \), is a triprotic acid, meaning it can donate three \( \mathrm{H}^+ \) ions. It reacts with potassium hydroxide (KOH), a strong base, which provides \( \mathrm{OH}^- \) ions. In the reaction, each \( \mathrm{H}^+ \) ion from citric acid reacts with one \( \mathrm{OH}^- \) ion to form water. Thus, the complete reaction involves three moles of KOH per mole of citric acid.
02

Write the Net Ionic Equation

The net ionic equation for the reaction between citric acid and KOH can be represented as follows:\[\mathrm{C}_6 \mathrm{H}_5 \mathrm{O}_7^{3-} + 3 \mathrm{OH}^- \rightarrow 3 \mathrm{H}_2 \mathrm{O} + \mathrm{C}_6 \mathrm{H}_5 \mathrm{O}_7^{3-}\]Note that \( \mathrm{C}_6 \mathrm{H}_5 \mathrm{O}_7^{3-} \) represents the fully deprotonated form of citric acid (citrate) after releasing all three protons.
03

Calculate Moles of KOH Used

To find the moles of KOH used, use the formula:\[\text{Moles of KOH} = \text{Volume (L)} \times \text{Molarity (M)}\]Substituting the given values:\[23.20 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} \times 0.500 \mathrm{M} = 0.0116 \text{ moles of KOH}\]
04

Relate Moles of KOH to Moles of Citric Acid

Since the reaction requires three moles of KOH to neutralize one mole of citric acid, we can set up the following relationship:\[\text{Moles of citric acid} = \frac{\text{Moles of KOH}}{3}\]Calculating gives:\[\frac{0.0116}{3} = 0.003867 \text{ moles of citric acid}\]
05

Calculate Molarity of Citric Acid Solution

The molarity (M) of the citric acid solution is calculated by dividing the moles of citric acid by the volume of the citric acid solution in liters:\[\text{Molarity of citric acid} = \frac{\text{Moles of citric acid}}{\text{Volume (L)}}\]Substitute the given volume of the citric acid solution:\[\frac{0.003867}{0.100} = 0.03867 \mathrm{M}\]
06

Conclusion

The balanced net ionic equation for the reaction is:\[ \mathrm{C}_6 \mathrm{H}_5 \mathrm{O}_7^{3-} + 3 \mathrm{OH}^- \rightarrow 3 \mathrm{H}_2 \mathrm{O} + \mathrm{C}_6 \mathrm{H}_5 \mathrm{O}_7^{3-} \]The molarity of the citric acid solution is \( 0.03867 \mathrm{M} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Balanced Chemical Equation
A balanced chemical equation is essential in chemical titrations to ensure stoichiometry is maintained, reflecting the precise relationship between reactants and products. In the case of a neutralization reaction involving citric acid (a triprotic acid) and potassium hydroxide (KOH, a strong base), accurately balancing the equation is crucial. This reaction involves citric acid releasing its three protons (H⁺ ions) to react with hydroxide ions (OH⁻), forming water (H₂O) and the conjugate citrate ion.
The balanced net ionic equation is:
  • \[ \mathrm{C}_6\mathrm{H}_5\mathrm{O}_7^{3-} + 3 \mathrm{OH}^- \rightarrow 3 \mathrm{H}_2\mathrm{O} + \mathrm{C}_6\mathrm{H}_5\mathrm{O}_7^{3-} \]
This balance ensures all three acidic protons of citric acid react fully with hydroxide ions. Balance is vital not just for writing the equation but also for quantifying the exact amount of reactants required, which is particularly important in laboratory settings.
Molarity Calculation
Calculating molarity is a key skill in understanding concentrations in solutions. Molarity (M) is defined as the number of moles of solute per liter of solution. In titrations, it helps determine unknown concentrations by using a titrant with known molarity.
For this exercise, the molarity of KOH used is known (0.500 M), which enables us to calculate the moles of KOH using:
  • \[ \text{Moles of KOH} = \text{Volume (L)} \times \text{Molarity (M)} \]
  • \[ 0.0116 \text{ moles of KOH, based on the volume of } 23.20 \text{ mL.} \]
From this, we related the moles of KOH to moles of citric acid, knowing the stoichiometric relationship (3:1). Finally, converting moles into molarity for citric acid by dividing by its solution volume (0.100 L) gives 0.03867 M. Understanding these calculations ensures precise knowledge of solution concentrations, a fundamental aspect of titration analysis.
Triprotic Acid
Triprotic acids, like citric acid in this exercise, contain three protons that can be donated, meaning they can undergo three stages of ionization. Each stage can release a proton (H⁺), and typically, the strength of ionization decreases with each stage.
In reactions with bases, these protons are sequentially neutralized, often requiring different amounts and strengths of base. In citric acid's case, it interacts with KOH such that each mole of citric acid is completely neutralized with three moles of KOH.
This characteristic of triprotic acids is crucial in analytical procedures like titration as it affects the stoichiometry of reactions occurring. Acknowledging the triprotic nature of citric acid ensures accurate calculations in determining concentrations and understanding of acid strength in various solutions.
Neutralization Reaction
A neutralization reaction occurs when an acid and a base react to form water and a salt, a fundamental type of chemical process. In the exercise's context, citric acid, a triprotic acid, and KOH, a strong base, undergo neutralization.
Each of the protons from the acidic solution reacts with a hydroxide ion from the base, resulting in water formation. The reaction can be expressed as a balanced net ionic equation, helping visualize how reactants transform into products.
Understanding this process emphasizes the stoichiometric relations where three moles of KOH fully neutralize one mole of citric acid. Neutralization not only highlights the chemical change but also the completion point of titration, called the equivalence point, which is fundamental in successfully performing and analyzing titrations.

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Most popular questions from this chapter

Uranium hexafluoride, \(\mathrm{UF}_{6},\) is processed to produce fuel for nuclear reactors and nuclear weapons. UF \(_{6}\) can be produced in a two-step reaction. Solid uranium (IV) oxide, \(\mathrm{UO}_{2}\), is first made to react with hydrofluoric acid (HF) solution to form solid UF \(_{4}\) with water as a by-product. UF \(_{4}\) further reacts with fluorine gas to form \(\mathrm{UF}_{6}\). (a) Write the balanced molecular equations for the conversion of \(\mathrm{UO}_{2}\) into \(\mathrm{UF}_{4}\) and the conversion of \(\mathrm{UF}_{4}\) to \(\mathrm{UF}_{6}\). (b) Which step is an acid-base reaction? (c) Which step is a redox reaction?

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