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Chapter 15: Problem 20

A buffer is prepared by dissolving \(\mathrm{HONH}_{2}\) and \(\mathrm{HONH}_{3} \mathrm{NO}_{3}\) in some water. Write equations to show how this buffer neutralizes added \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\).

Short Answer

Expert verified
In conclusion, the buffer solution containing HONH鈧 and HONH鈧僋O鈧 can neutralize added H鈦 and OH鈦 ions. The equilibrium reactions for these neutralization processes are: \(HONH_2 + H^+ \rightleftharpoons HONH_3^+\) (for added H鈦 ions) \(HONH_3NO_3 \rightleftharpoons H^+ + NO_3^- + HONH_2\) (for added OH鈦 ions)

Step by step solution

01

1. Identify the reactions for the buffer components

: In a buffer solution containing HONH鈧 and HONH鈧僋O鈧, the following acid-base reactions occur: - HONH鈧 can accept an H鈦 ion and form its conjugate acid HONH鈧冣伜 - HONH鈧僋O鈧 can donate an H鈦 ion and form its conjugate base HONH鈧 These reactions allow the buffer to neutralize added H鈦 and OH鈦 ions.
02

2. Write the equilibrium reaction for the weak acid HONH鈧

: For the weak base HONH鈧 accepting an H鈦 ion, the equilibrium reaction is: \(HONH_2 + H^+ \rightleftharpoons HONH_3^+\) Here, the HONH鈧 is in equilibrium with its conjugate acid, HONH鈧冣伜. When an H鈦 ion is added to the solution, the equilibrium will shift to the right, and more HONH鈧冣伜 will be formed, neutralizing the added H鈦 ion.
03

3. Write the equilibrium reaction for the weak acid HONH鈧僋O鈧

: For the weak acid HONH鈧僋O鈧 donating an H鈦 ion, the equilibrium reaction is: \(HONH_3NO_3 \rightleftharpoons H^+ + NO_3^- + HONH_2\) Here, HONH鈧僋O鈧 is in equilibrium with its conjugate base, HONH鈧. When an OH鈦 ion is added to the solution, it reacts with the H鈦 ion in the solution, reducing its concentration. The equilibrium will shift to the right, and more HONH鈧 and H鈦 ions will be formed. The newly formed H鈦 ions will neutralize the added OH鈦 ions. In conclusion, the buffer solution containing HONH鈧 and HONH鈧僋O鈧 can neutralize added H鈦 and OH鈦 ions, maintaining the pH of the solution. The equilibrium reactions for these neutralization processes are: \(HONH_2 + H^+ \rightleftharpoons HONH_3^+\) (for added H鈦 ions) \(HONH_3NO_3 \rightleftharpoons H^+ + NO_3^- + HONH_2\) (for added OH鈦 ions)

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Key Concepts

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

Understanding Acid-Base Equilibrium
In buffer solutions, acid-base equilibrium is essential for maintaining a stable pH. The system is composed of weak acids and their conjugate bases. They can react with added acids or bases to minimize changes in pH.

In our buffer, HONH鈧 acts as a weak base and can accept additional protons (H鈦 ions) to become the conjugate acid HONH鈧冣伜. This equilibrium is expressed as:
  • \(HONH_2 + H^+ \rightleftharpoons HONH_3^+\)
When extra H鈦 is introduced into the solution, the equilibrium shifts right. This means more HONH鈧冣伜 is created to neutralize the increase in acidity. Conversely, if OH鈦 ions are added, they react with H鈦 to form water, reducing H鈦 concentration and shifting equilibrium to the left.

Understanding this dynamic process helps in predicting how different components affect the pH of a solution.
Conjugate Acid-Base Pairs Explained
A conjugate acid-base pair consists of two species that transform into each other by gain or loss of a proton. In our buffer scenario, HONH鈧 and HONH鈧冣伜 form such a pair.

  • HONH鈧 is a weak base because it can accept an H鈦 ion, forming HONH鈧冣伜.
  • HONH鈧冣伜 is the conjugate acid of HONH鈧. It can donate an H鈦 to revert back to HONH鈧.
These pairs are critical in buffering systems. They allow the solution to resist drastic pH changes. When H鈦 is added, the base component (HONH鈧) of the pair reacts with it. When OH鈦 is added, the acid component (HONH鈧冣伜) reacts, donating a proton to form water.

This balance through conjugate pairs is what permits buffers to stabilize pH efficiently.
How Neutralization Reactions Work in Buffers
Neutralization reactions occur when acids and bases react to form water and salt, reducing acidity or basicity. In buffers, these reactions help maintain a stable pH environment.

When we add H鈦 ions to the buffer, HONH鈧 captures them. The equation shows it transforming into the conjugate acid HONH鈧冣伜:
  • \(HONH_2 + H^+ \rightleftharpoons HONH_3^+\)
Adding OH鈦 induces reaction with available H鈦 to form water:
  • \(OH^- + H^+ \rightarrow H_2O\)
  • Excess HONH鈧 is converted effectively, maintaining pH.
The buffer moderates pH changes by utilizing these reactions.

By understanding how neutralization reactions are used, we see how buffers work as protective systems in chemical processes.

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Most popular questions from this chapter

Sketch a pH curve for the titration of a weak acid (HA) with a strong base (NaOH). List the major species, and explain how you would go about calculating the \(\mathrm{pH}\) of the solution at various points, including the halfway point and the equivalence point.

What concentration of \(\mathrm{NH}_{4} \mathrm{Cl}\) is necessary to buffer a \(0.52-M\) \(\mathrm{NH}_{3}\) solution at \(\mathrm{pH}=9.00 ?\left(K_{\mathrm{b}}\right.\) for \(\left.\mathrm{NH}_{3}=1.8 \times 10^{-5} .\right)\)

A friend asks the following: "Consider a buffered solution made up of the weak acid HA and its salt NaA. If a strong base like \(\mathrm{NaOH}\) is added, the HA reacts with the \(\mathrm{OH}^{-}\) to form \(\mathrm{A}^{-}\). Thus the amount of acid (HA) is decreased, and the amount of base \(\left(\mathrm{A}^{-}\right)\) is increased. Analogously, adding \(\mathrm{HCl}\) to the buffered solution forms more of the acid (HA) by reacting with the base \(\left(\mathrm{A}^{-}\right) .\) Thus how can we claim that a buffered solution resists changes in the \(\mathrm{pH}\) of the solution?" How would you explain buffering to this friend?

What are the major species in solution after \(\mathrm{NaHSO}_{4}\) is dissolved in water? What happens to the \(\mathrm{pH}\) of the solution as more \(\mathrm{NaHSO}_{4}\) is added? Why? Would the results vary if baking soda \(\left(\mathrm{NaHCO}_{3}\right)\) were used instead?

Consider the titration of \(40.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HClO}_{4}\) by \(0.100 M \mathrm{KOH}\). Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of KOH have been added. a. \(0.0 \mathrm{~mL}\) b. \(10.0 \mathrm{~mL}\) c. \(40.0 \mathrm{~mL}\) d. \(80.0 \mathrm{~mL}\) e. \(100.0 \mathrm{~mL}\)
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