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Explain why a mixture formed by mixing \(100 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) and \(50 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\) will act as a buffer.

Short Answer

Expert verified
The mixture formed by mixing 100 mL of 0.100 M CH3COOH and 50 mL of 0.100 M NaOH will act as a buffer because after the reaction between the weak acid (CH3COOH) and the strong base (NaOH), both the weak acid and its conjugate base (CH3COO-) remain in the solution, with 0.005 mol of each present. This allows the mixture to resist significant changes in pH when small amounts of acid or base are added, as the CH3COOH can neutralize added base and the CH3COO- can neutralize added acid, maintaining a relatively constant pH.

Step by step solution

01

Calculate the moles of CH3COOH and NaOH in the solution

First, we need to find out the amount of moles of the acetic acid and sodium hydroxide present in the given volumes of the solutions. We can use the formula: moles = Molarity × Volume For CH3COOH: moles = 0.100 M × 100 mL = 0.100 mol/L × 0.100 L = 0.010 mol For NaOH: moles = 0.100 M × 50 mL = 0.100 mol/L × 0.050 L = 0.005 mol
02

Determine the reaction between CH3COOH and NaOH

Next, we need to determine the reaction that takes place when the acetic acid and sodium hydroxide are mixed. The reaction between a weak acid and a strong base is as follows: CH3COOH + NaOH → CH3COONa + H2O The sodium acetate (CH3COONa) will dissociate completely into its ions and will have no effect on the pH of the solution. The important species remaining in the solution are CH3COOH and its conjugate base CH3COO- formed as a result of the reaction.
03

Calculate the moles of CH3COOH and CH3COO- after reaction

Now, we need to calculate the moles of CH3COOH and CH3COO- remaining in the solution after the reaction between CH3COOH and NaOH is complete. Since the reaction goes to completion, the limiting reagent determines the extent of the reaction. The moles of CH3COOH and NaOH are 0.010 and 0.005, respectively. As NaOH is the limiting reagent, the reaction proceeds as follows: 0.010 mol CH3COOH - 0.005 mol NaOH → 0.005 mol CH3COOH + 0.005 mol CH3COO- Thus, after the reaction, there are 0.005 mol of CH3COOH and 0.005 mol of CH3COO- in the solution.
04

Explain why the mixture acts as a buffer

As we have determined that there are both the weak acid (CH3COOH) and its conjugate base (CH3COO-) present in the solution after the reaction, the mixture can act as a buffer. The acetic acid and acetate ion help to maintain the pH of the solution by neutralizing any additional acid or base added to the mixture. The acetic acid can react with an added base, while the acetate ion can react with an added acid. This will keep the pH of the solution relatively constant and confirms that the given mixture acts as a buffer.

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Key Concepts

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

Acetic Acid
Acetic acid is a simple, yet essential component in understanding buffer solutions. It’s often represented by the chemical formula \( \text{CH}_3\text{COOH} \). As a weak acid, acetic acid doesn't fully ionize in water. This means that when dissolved, it yields relatively few hydrogen ions compared to a strong acid.
It’s these incomplete ionizations that make acetic acid an ideal candidate for creating buffer solutions. When mixed with a strong base, such as sodium hydroxide, it reacts to form a mixture that can resist changes in pH.
  • Properties: Acetic acid is a colorless liquid with a vinegar-like odor.
  • Use in buffers: It maintains stability in solutions by using its acidic properties.
  • Ionization: Ionizes partially, contributing to its role in buffer systems.
Understanding acetic acid's role is crucial when examining buffer solutions as it provides the acidic component necessary for buffering action.
Conjugate Base
The concept of a conjugate base emerges from the interaction between a weak acid and a base. When acetic acid reacts with sodium hydroxide, it forms acetate ions \( \text{CH}_3\text{COO}^- \), which is its conjugate base.
The conjugate base is significant because it can react with added acids to prevent large swings in pH, playing a vital role in the buffer system.
  • Formation: Result of a weak acid losing a hydrogen ion.
  • Function in buffers: Helps neutralize added acids, maintaining pH balance.
  • Behavior: Reacts readily with hydrogen ions in solution.
In a buffer solution, the presence of both the weak acid and its conjugate base is what enables the solution to maintain its pH despite the addition of external acids or bases.
Weak Acid
Weak acids, like acetic acid, are distinct in their incomplete ionization in water. This characteristic allows them to partake in buffering systems by both donating and accepting hydrogen ions.
Compared to strong acids, weak acids only partially disassociate into ions in solution. This allows them to participate in reversible reactions with bases.
  • Ionic dissociation: Releases fewer hydrogen ions.
  • Buffer role: Interacts with bases to stabilize pH levels.
  • Reversibility: Can reclaim lost hydrogen ions, an essential part of buffering capabilities.
Understanding weak acids is fundamental in chemistry, as they provide the balance required to maintain the pH of solutions in buffer systems.
Strong Base
A strong base, such as sodium hydroxide (NaOH), fully dissociates in water providing a high concentration of hydroxide ions \( \text{OH}^- \).
The interaction of a strong base with a weak acid forms the basis of a buffer solution, as it facilitates the creation of a conjugate base.
  • Dissociation: Completely ionizes in an aqueous solution.
  • Reaction with acids: Can neutralize acids effectively.
  • Role in buffering: Helps in forming the conjugate base critical for a buffer system.
In our context with acetic acid, the strong base reacts just enough to convert some of it into its conjugate base, balancing the components necessary for buffering action.
pH Stability
The ability of a buffer solution to resist changes in pH is known as pH stability. This is crucial in many chemical and biological processes where maintaining a consistent pH is necessary.
In a buffer system like the one made from acetic acid and its conjugate base, the solution can adjust by neutralizing added acids or bases, ensuring that the pH remains relatively constant.
  • Mechanism: Acid components neutralize bases, while conjugate bases neutralize added acids.
  • Application: Used in various applications, from chemical reactions to biological systems, wherever pH control is essential.
  • Efficiency: Highly dependent on the relative concentrations of the acid and its conjugate base.
Mastery of pH stability and buffering is fundamental in fields such as chemistry, biology, and environmental science, where small pH changes could have significant implications.

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Most popular questions from this chapter

Aspirin has the structural formula At body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{a}\) for aspirin equals \(3 \times 10^{-5}\). If two aspirin tablets, each having a mass of \(325 \mathrm{mg}\), are dissolved in a full stomach whose volume is \(1 \mathrm{~L}\) and whose \(\mathrm{pH}\) is 2, what percent of the aspirin is in the form of neutral molecules?

Furoic acid \(\left(\mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) has a \(K_{a}\) value of \(6.76 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\). Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of \((\) a) a solution formed by adding \(25.0 \mathrm{~g}\) of furoic acid and \(30.0 \mathrm{~g}\) of sodium furoate \(\left(\mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) to enough water to form \(0.250 \mathrm{~L}\) of solution; (b) a solution formed by mixing \(30.0 \mathrm{~mL}\) of \(0.250 \mathrm{MHC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\) and \(20.0 \mathrm{~mL}\) of \(0.22 \mathrm{M} \mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\) and diluting the total volume to \(125 \mathrm{~mL} ;\) (c) a solution prepared by adding \(50.0 \mathrm{~mL}\) of \(1.65 \mathrm{M} \mathrm{NaOH}\) solution to \(0.500\) Lof \(0.0850 \mathrm{M} \mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\).

A solution containing an unknown number of metal ions is treated with dilute \(\mathrm{HCl}\); no precipitate forms. The \(\mathrm{pH}\) is adjusted to about 1 , and \(\mathrm{H}_{2} \mathrm{~S}\) is bubbled through. Again, no precipitate forms. The pH of the solution is then adjusted to about 8 . Again, \(\mathrm{H}_{2} \mathrm{~S}\) is bubbled through. This time a precipitate forms. The filtrate from this solution is treated with \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\). No precipitate forms.

Equal quantities of \(0.010 \mathrm{M}\) solutions of an acid \(\mathrm{HA}\) and a base \(B\) are mixed. The \(p H\) of the resulting solution is 9.2. (a) Write the equilibrium equation and equilibriumconstant expression for the reaction between HA and B. (b) If \(K_{a}\) for \(\mathrm{HA}\) is \(8.0 \times 10^{-5}\), what is the value of the equilibrium constant for the reaction between HA and B? (c) What is the value of \(K_{b}\) for B?

To what final concentration of \(\mathrm{NH}_{3}\) must a solution be adjusted to just dissolve \(0.020 \mathrm{~mol}\) of \(\mathrm{NiC}_{2} \mathrm{O}_{4}\) \(\left(K_{s p}=4 \times 10^{-10}\right)\) in \(1.0 \mathrm{~L}\) of solution? (Hint: You can neglect the hydrolysis of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) because the solution will be quite basic.)

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