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Which of the following can be classified as buffer solutions? a. \(0.25 \mathrm{M} \mathrm{HBr}+0.25 \mathrm{M} \mathrm{HOBr}\) b. \(0.15 \mathrm{M} \mathrm{HClO}_{4}+0.20 \mathrm{M} \mathrm{RbOH}\) c. \(0.50 \mathrm{M} \mathrm{HOCl}+0.35 \mathrm{M} \mathrm{KOCl}\) d. \(0.70 \mathrm{MKOH}+0.70 \mathrm{M} \mathrm{HONH}_{2}\) e. \(0.85 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}+0.60 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{3} \mathrm{NO}_{3}\)

Short Answer

Expert verified
Option c: \(0.50\,\mathrm{M}\,\mathrm{HOCl} + 0.35\,\mathrm{M}\,\mathrm{KOCl}\) and option e: \(0.85\,\mathrm{M}\,\mathrm{H}_2\,\mathrm{NNH}_2 + 0.60\,\mathrm{M}\, \mathrm{H}_2\,\mathrm{NNH}_3\,\mathrm{NO}_3\) can be classified as buffer solutions.

Step by step solution

01

Option a: \(0.25\,\mathrm{M}\,\mathrm{HBr} + 0.25\,\mathrm{M}\,\mathrm{HOBr}\)

Here, HBr is a strong acid, and HOBr is a weak acid. Since the solution contains a strong acid, it cannot be a buffer solution, as the presence of strong acids makes the solution unable to resist pH changes.
02

Option b: \(0.15\,\mathrm{M}\,\mathrm{HClO}_4 + 0.20\,\mathrm{M}\,\mathrm{RbOH}\)

HClO4 is a strong acid, and RbOH is a strong base. The presence of both a strong acid and strong base will neutralize the solution, but it will not create a buffer solution.
03

Option c: \(0.50\,\mathrm{M}\,\mathrm{HOCl} + 0.35\,\mathrm{M}\,\mathrm{KOCl}\)

This solution contains HOCl (a weak acid) and its conjugate base OCl- (from KOCl). Both the weak acid and its conjugate base are present in significant amounts in the solution, making this a buffer solution.
04

Option d: \(0.70\,\mathrm{M}\,\mathrm{KOH} + 0.70\,\mathrm{M}\,\mathrm{HONH}_2\)

In this case, KOH is a strong base, and HONH2 is a weak acid. As we have a strong base present, the solution will not be able to resist changes in pH, so it is not a buffer solution.
05

Option e: \(0.85\,\mathrm{M}\,\mathrm{H}_2\,\mathrm{NNH}_2 + 0.60\,\mathrm{M}\, \mathrm{H}_2\,\mathrm{NNH}_3\,\mathrm{NO}_3\)

Here, \(H_2NNH_2\) (hydrazine) is a weak base, and its conjugate acid is \(\mathrm{H}_{2} \mathrm{NNH}_{3}^{+}\) which is present in the form of its salt \(\mathrm{H}_{2}\mathrm{NNH}_{3}\mathrm{NO}_{3}\) in the solution. This system consists of a weak base and its conjugate acid in significant amounts, so it can be classified as a buffer solution. In conclusion, option c and option e can be classified as buffer solutions.

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Key Concepts

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

Weak Acid and Conjugate Base
A buffer solution is typically formed by combining a weak acid and its conjugate base. This duo works together to stabilize pH levels. A weak acid, such as HOCl, only partially ionizes in water. This means it doesn't release all of its hydrogen ions.
  • For instance, in the reaction of HOCl and KOCl, HOCl donates protons, but its conjugate base, OCl-, can accept them.
  • This reversible reaction allows the system to maintain a relatively stable pH, even when a small amount of acid or base is added.
Effective buffers contain significant amounts of both the weak acid and its conjugate base. They must exist in comparable concentrations to efficiently absorb excess H+ or OH- ions.
Understanding the behavior of weak acids and their conjugate bases is crucial in predicting whether a mixture can act as a buffer solution.
Weak Base and Conjugate Acid
A buffer solution can also be prepared by using a weak base and its conjugate acid. Unlike strong bases, weak bases only partially accept protons in water. This makes their pairing with conjugate acids ideal for pH buffering functions.
In the example provided, the combination of hydrazine ( H_2NNH_2 ) and its conjugate acid H_2NNH_3 NO_3 serves as an excellent buffer system.
  • The weak base temporarily accepts a proton, transforming into its conjugate acid.
  • Conversely, when the buffer encounters excess base, the conjugate acid donates a proton, reverting back to the weak base form.
This seesaw battle between the acid and base forms helps keep the pH relatively stable. These properties make weak base-conjugate acid pairs another powerful strategy in creating buffer solutions.
pH Resistance
Buffers are specifically designed to resist changes in pH upon addition of small amounts of acids or bases. Their pH resistance is a measure of how well they can hold their ground against fluctuations. This is possible due to the equilibrium between the weak acid/base and their conjugates.
Imagine the buffer solution as a sponge. When you add small amounts of acid or base, this sponge can absorb or release ions.
  • Adding acid increases H+ ions, but the buffer responds by converting base to its conjugate acid.
  • On the flip side, adding a base decreases H+ ions, leading the buffer to convert its acid to the conjugate base.
This dynamic process is what gives buffers their ability to regulate pH within a certain range, typically close to the pKa of the weak acid or weak base used.
Strong Acid and Strong Base
When strong acids or bases are introduced into a solution, they dissociate completely. This is why they aren't suitable for creating buffer solutions. The complete ionization results in a one-way conversion that fails to provide any pH resistance.
Consider the presence of either a strong acid like HClO4 or a strong base like RbOH.
  • In such cases, the strong component reacts drastically with the buffer system, leading to a significant change in pH.
  • Neither species offers the reversible reaction needed to establish a stable pH level, since the strong acids/bases ionize completely.
Thus, the presence of any strong acid or base in the solutions from the exercise ruled out their capability of being buffers. The complete dissociation disrupts the balanced equilibrium necessary for pH buffering.

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Most popular questions from this chapter

What volumes of \(0.50 \mathrm{M} \mathrm{HNO}_{2}\) and \(0.50 \mathrm{M} \mathrm{NaNO}_{2}\) must be mixed to prepare \(1.00 \mathrm{~L}\) of a solution buffered at \(\mathrm{pH}=3.55\) ?

A buffered solution is made by adding \(50.0 \mathrm{~g} \mathrm{NH}_{4} \mathrm{Cl}\) to \(1.00\) \(\mathrm{L}\) of a \(0.75-M\) solution of \(\mathrm{NH}_{3} .\) Calculate the \(\mathrm{pH}\) of the final solution. (Assume no volume change.)

Consider the titration of \(100.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}\) \(\left(K_{\mathrm{b}}=3.0 \times 10^{-6}\right)\) by \(0.200 \mathrm{M} \mathrm{HNO}_{3} .\) Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of \(\mathrm{HNO}_{3}\) have been added. a. \(0.0 \mathrm{~mL}\) b. \(20.0 \mathrm{~mL}\) c. \(25.0 \mathrm{~mL}\) d. \(40.0 \mathrm{~mL}\) e. \(50.0 \mathrm{~mL}\) f. \(100.0 \mathrm{~mL}\)

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?

Calculate the \(\mathrm{pH}\) at the halfway point and at the equivalence point for each of the following titrations. a. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) titrated by \(0.10 \mathrm{M} \mathrm{NaOH}\) b. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\) titrated by \(0.20 \mathrm{M} \mathrm{HNO}_{3}\) c. \(100.0 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HCl}\) titrated by \(0.25 \mathrm{M} \mathrm{NaOH}\)

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