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Could a buffered solution be made by mixing aqueous solutions of \(\mathrm{HCl}\) and \(\mathrm{NaOH}\) ? Explain. Why isn't a mixture of a strong acid and its conjugate base considered a buffered solution?

Short Answer

Expert verified
A buffered solution cannot be made by mixing aqueous solutions of HCl and NaOH, as they undergo a neutralization reaction to form water and NaCl without forming a proper conjugate system. The solution lacks buffering capacity, which is required to resist changes in pH upon the addition of small amounts of acid or base. This is why a mixture of a strong acid and its conjugate base is not considered a buffered solution.

Step by step solution

01

Identify the nature of the given compounds

HCl is a strong acid, and NaOH is a strong base. When these two compounds are mixed in an aqueous solution, they will undergo complete ionization.
02

Chemical reaction between HCl and NaOH

When HCl and NaOH are mixed in an aqueous solution, they will react together to form water and a salt (NaCl) through a neutralization reaction. The balanced chemical equation for this reaction is: \[ \mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{H_{2}O} + \mathrm{NaCl} \]
03

Determine the conjugate system

In the case of a strong acid and a strong base, there is no proper conjugate system formed. The reaction between HCl and NaOH produces water, in which the H+ ions from HCl and the OH- ions from NaOH combine to form neutral water molecules, not creating any additional buffer ions or molecules. The salt, NaCl, is not involved in any further reactions that can help maintain the pH of the solution.
04

Buffering capacity

Buffer solutions should be able to maintain their pH upon the addition of small amounts of acid or base. In the case of a mixture of HCl and NaOH, the solution has no buffering capacity, as there is no weak acid or weak base present, which typically shows resistance to changes in pH when small amounts of acid or base are added.
05

Conclusion

A buffered solution cannot be made by mixing aqueous solutions of HCl and NaOH, as they completely react with each other in a neutralization reaction with no proper conjugate system formed. A mixture of a strong acid and its conjugate base is not considered a buffered solution because it lacks the buffering capacity needed to resist changes in pH upon the addition of small amounts of acid or base.

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Key Concepts

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

Strong Acids
Strong acids, like hydrochloric acid (HCl), are substances that completely dissociate into their ions when dissolved in water. This means that in a solution, the strong acid releases all its hydrogen ions (H+) into the water. As a result, these solutions have a very low pH.

Key characteristics of strong acids include:
  • Complete ionization in solution, meaning they break down fully into their ions.
  • Their ability to conduct electricity well, due to the high concentration of ions.
  • Inability to form buffers because they do not leave behind any reserve of the acid to neutralize added bases.
This complete ionization is why in the mixture of HCl and NaOH, we do not see any buffering capacity, as opposed to mixtures with weak acids that only partially dissociate.
Neutralization Reaction
A neutralization reaction occurs when an acid and a base react to form water and a salt. In the case of HCl and NaOH mixing, they undergo a classic neutralization reaction:

\[ \mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{H_2O} + \mathrm{NaCl} \]Here, the hydrogen ion (H+) from HCl meets the hydroxide ion (OH−) from NaOH, creating water (H2O). The remaining ions, Na+ and Cl-, combine to form sodium chloride (NaCl), a common salt.

Neutralization reactions:
  • Are typically exothermic, releasing energy as heat.
  • Result in a tangible change in the properties of a solution, often evidenced by a shift in pH towards neutrality (pH 7).
  • Are not capable of creating a buffer because the reactants are fully consumed, leaving no residual acid or base to resist pH changes.
Buffering Capacity
Buffering capacity refers to the ability of a solution to resist changes in pH upon the addition of small amounts of acid or base. A good buffer solution contains:

  • A weak acid and its conjugate base, or
  • A weak base and its conjugate acid.
These components exist in equilibrium, allowing the solution to absorb excess H+ or OH- without a significant change in the pH level.

In the context of HCl and NaOH, the complete reaction leaves no room for buffering, as no weak acid or base pairs are present. Instead of buffering, such a combination results in a neutral solution of water and salt, which cannot resist pH changes effectively.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products. Key aspects of such reactions include the rearrangement of atoms and changes in energy.

For a reaction like the one between HCl and NaOH:
  • The rearrangement of ions occurs, where H+ from HCl pairs with OH- from NaOH to form water.
  • This reaction showcases an energetically favorable transformation, releasing energy and leading to more stable products.
  • Often follows a specific pattern or pathway, like neutralization in this scenario.
Such reactions are foundational to understanding how certain properties of solutions, such as pH, change in chemistry. By learning how to predict the outcomes of these reactions, we form a clearer picture of how chemical species interact.

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Most popular questions from this chapter

Phosphate buffers are important in regulating the \(\mathrm{pH}\) of intracellular fluids at \(\mathrm{pH}\) values generally between \(7.1\) and \(7.2\). a. What is the concentration ratio of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) to \(\mathrm{HPO}_{4}{ }^{2-}\) in intracellular fluid at \(\mathrm{pH}=7.15\) ? \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \rightleftharpoons \mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=6.2 \times 10^{-8}\) b. Why is a buffer composed of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ineffective in buffering the \(\mathrm{pH}\) of intracellular fluid? \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=7.5 \times 10^{-3}\)

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?

Consider the titration of \(100.0 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) acetic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\) 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. \(50.0 \mathrm{~mL}\) c. \(100.0 \mathrm{~mL}\) d. \(150.0 \mathrm{~mL}\) e. \(200.0 \mathrm{~mL}\) f. \(250.0 \mathrm{~mL}\)

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. \(100.0 \mathrm{~mL}\)

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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