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A buffer is prepared by dissolving HONH_ 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
The buffer solution neutralizes added H+ and OH- ions through the following reactions: \[HONH_2 + H^+ \rightarrow HONH_3^+\] \[HONH_3^+ + OH^- \rightarrow HONH_2 + H_2O\]

Step by step solution

01

Write the ionization reaction for HONH3+ and the reaction for HONH2 with water

First, let us write down the ionization reaction for HONH3+ (conjugate acid) and the reaction for HONH2 (conjugate base) with water: \[HONH_3^+ + H_2O \rightleftharpoons HONH_2 + H_3O^+\] \[HONH_2 + H_2O \rightleftharpoons HONH_3^+ + OH^-\]
02

Determine how these reactions help buffer the solution when H+ and OH- ions are added

When H+ ions are added to the buffer solution, the following reaction will occur: \[HONH_2 + H^+ \rightarrow HONH_3^+\] This reaction shows that the buffer will neutralize the added H+ ions by using the conjugate base (HONH2) to form the conjugate acid (HONH3+), thereby preventing any significant change in the pH of the solution. Now let us see how the buffer neutralizes added OH- ions: \[HONH_3^+ + OH^- \rightarrow HONH_2 + H_2O\] In this case, the buffer uses the conjugate acid (HONH3+) to react with the added OH- ions, forming the conjugate base (HONH2) and water. This process neutralizes the added OH- ions and prevents significant pH changes in the solution. So, the two equations that show how this buffer solution neutralizes added H+ and OH- ions are: \[HONH_2 + H^+ \rightarrow HONH_3^+\] \[HONH_3^+ + OH^- \rightarrow HONH_2 + H_2O\]

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Key Concepts

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

Acid-Base Equilibrium
In chemistry, acid-base equilibrium refers to the balance between acidic and basic molecules in a solution. This equilibrium is crucial in understanding how solutions will react when external acids or bases are added.
Acid-base equilibrium involves reversible reactions. These reactions can proceed in both forward and backward directions.
For example, in the buffer solution containing HONH_3^+ and HONH_2, an equilibrium exists between these molecules and their associated ions:
  • HONH_3^+ dissociates into HONH_2 and H_3O^+.
  • HONH_2 reacts with water to form HONH_3^+ and OH^-.
Manipulation of this equilibrium occurs when additional H^+ or OH^- is introduced to the system. The solution adjusts to maintain equilibrium, thus resisting significant pH changes. Understanding this dynamic interaction is key to grasping how buffer solutions function.
Conjugate Acid-Base Pair
The concept of a conjugate acid-base pair is fundamental in acid-base chemistry. It refers to pairs of compounds that transform into each other by gaining or losing a proton (H^+).
In our buffer example, HONH_3^+ and HONH_2 form a conjugate acid-base pair.
  • HONH_3^+ is the conjugate acid, capable of donating a proton to become HONH_2.
  • HONH_2 is the conjugate base, accepting a proton to become HONH_3^+.
This reversible transfer of protons allows the buffer to neutralize added acids or bases effectively. By utilizing the conjugate pairs, the system can absorb increases in H^+ or OH^- without a drastic change in pH. This seamless proton exchange is what gives buffer solutions their stability in maintaining constant pH levels.
Buffer Solution Mechanism
Buffer solutions are specially formulated to resist significant changes in pH upon the addition of small amounts of acid or base. The mechanism behind this involves the use of a weak acid and its conjugate base (or vice versa).
When H^+ ions are introduced into the buffer solution, they interact with the conjugate base (HONH_2), converting it into the conjugate acid (HONH_3^+). This neutralization prevents an increase in acid concentration and thus stabilizes the pH.
  • HONH_2 + H^+ → HONH_3^+
Similarly, when OH^- ions are added, the conjugate acid (HONH_3^+) reacts with them, forming water and the conjugate base (HONH_2), counteracting the increase in pH caused by the added base.
  • HONH_3^+ + OH^- → HONH_2 + H_2O
Through this mechanism, buffer solutions maintain a stable pH, crucial in many chemical and biological processes. Understanding this concept allows us to utilize buffers effectively in various scientific applications.

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Most popular questions from this chapter

Calculate the pH of a solution prepared by mixing \(250 . \mathrm{mL}\) of 0.174 \(\mathrm{m}\) aqueous \(\mathrm{HF}\) (density \(=1.10 \mathrm{g} / \mathrm{mL} )\) with 38.7 \(\mathrm{g}\) of an aqueous solution that is 1.50\(\% \mathrm{NaOH}\) by mass (density \(=\) 1.02 \(\mathrm{g} / \mathrm{mL} ) .\left(K_{\mathrm{a}} \text { for } \mathrm{HF}=7.2 \times 10^{-4} .\right)\)

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 NaOH is added, the HA reacts with the OH2 to form A2. Thus the amount of acid (HA) is decreased, and the amount of base (A2) is increased. Analogously, adding HCl to the buffered solution forms more of the acid (HA) by reacting with the base (A2). Thus how can we claim that a buffered solution resists changes in the pH of the solution?” How would you explain buffering to this friend?

Repeat the procedure in Exercise \(67,\) but for the titration of 25.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{NH}_{3}\left(K_{\mathrm{b}}=1.8 \times 10^{-5}\right)\) with 0.100 \(\mathrm{M}\) \(\mathrm{HCl} .\)

Consider a solution formed by mixing 50.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}, 30.0 \mathrm{mL}\) of 0.100 \(\mathrm{M}\) HOCl, 25.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M}\) \(\mathrm{NaOH}, 25.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2},\) and 10.0 \(\mathrm{mL}\) of 0.150 \(M \mathrm{KOH}\) . Calculate the \(\mathrm{pH}\) of this solution.

Th pH of blood is steady at a value of approximately 7.4 as a result of the following equilibrium reactions: $$ \mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \leftrightharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \leftrightharpoons \mathrm{HCO}_{3}-(a q)+\mathrm{H}^{+}(a q) $$ The actual buffer system in blood is made up of \(\mathrm{H}_{2} \mathrm{CO}_{3}\) and \(\mathrm{HCO}_{3}\) - One way the body keeps the pH of blood at 7.4 is by regulating breathing. Under what blood ph conditions will the body increase breathing and under what blood pH conditions will the body decrease breathing? Explain.

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