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A best buffer has about equal quantities of weak acid and conjugate base present as well as having a large concentration of each species present. Explain.

Short Answer

Expert verified
A good buffer solution has equal quantities of weak acid and its conjugate base, along with high concentrations of each species, allowing it to effectively resist changes in pH when small amounts of acid or base are added. This is due to the Henderson-Hasselbalch equation, which shows that pH remains close to the weak acid's pKa when equal concentrations of weak acid and its conjugate base are present. The high concentrations of both the weak acid and its conjugate base ensure a high buffering capacity, enabling the buffer to neutralize larger amounts of acid or base while maintaining pH stability.

Step by step solution

01

Understanding Buffer Solutions

A buffer solution is an aqueous solution that resists changes in pH when small amounts of an acid or a base are added. This resistance is due to the presence of a weak acid and its conjugate base (or a weak base and its conjugate acid) that work together to neutralize added acid or base, keeping the pH relatively stable.
02

The Henderson-Hasselbalch Equation

The relationship between the pH of a buffer, the pKa of the weak acid, and the concentrations of the weak acid and its conjugate base can be described by the Henderson-Hasselbalch equation: \[pH = pKa + log\frac{[A^-]}{[HA]}\] where: - \(pH\) is the pH of the buffer - \(pKa\) is the negative logarithm of the acid dissociation constant, \(K_a\), of the weak acid - \([A^-]\) is the concentration of the conjugate base - \([HA]\) is the concentration of the weak acid
03

Importance of Equal Acid and Base Concentrations

According to the Henderson-Hasselbalch equation, when the concentrations of weak acid and conjugate base are equal (\([HA] = [A^-]\)), the \(pH\) of the solution is equal to the \(pKa\) of the weak acid: \[pH = pKa\] This means that the pH of the buffer solution remains close to the pKa value of the weak acid when equal concentrations of the weak acid and its conjugate base are present. This allows the buffer to effectively neutralize small amounts of added acid or base, preventing large pH changes.
04

Importance of High Concentrations of Acid and Base

Having a large concentration of both the weak acid and its conjugate base in the buffer ensures that the buffer has a high buffering capacity, which means it can neutralize a larger amount of acid or base added to it. When the concentration of the weak acid and its conjugate base is high, the pH will remain relatively stable despite the addition of relatively larger quantities of an acid or a base. This is because the buffer can absorb these changes without causing much change to the pH. In conclusion, a good buffer solution has approximately equal quantities of weak acid and conjugate base, as well as large concentrations of each species, to effectively resist changes in pH when small amounts of acid or base are added. This is explained by the Henderson-Hasselbalch equation and the concept of buffering capacity.

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Key Concepts

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

Henderson-Hasselbalch Equation
Understanding the Henderson-Hasselbalch equation is crucial when dealing with buffer solutions in chemistry. At its core, this equation provides a direct relationship between the pH of a buffer solution and the proportions of its constituent weak acid and conjugate base. The formula is expressed as:

\[ pH = pKa + \log\frac{[A^-]}{[HA]} \]

This simple equation tells us that by knowing the pKa of the weak acid in the buffer and the ratio of the concentrations of the conjugate base \([A^-]\) to the weak acid \([HA]\), we can calculate the pH of the buffer solution. From a practical standpoint, for students grappling with this equation, remember that the pKa is the pH at which equal amounts of the acid and its conjugate base are present. When these concentrations are equal, the logarithmic term effectively becomes zero, simplifying the equation to \( pH = pKa \). This provides a valuable insight into the desired state of a buffer solution where it functions most effectively—its pH is buffered at the pKa of the weak acid. For deeper comprehension, the logarithmic relationship implies that large changes in the buffer components' ratios will result in only modest pH changes, demonstrating the nature of the buffer's resiliency against pH fluctuations.
Acid-Base Neutralization
Acid-base neutralization is an essential process with direct implications on how buffer solutions manage to maintain a consistent pH level. Neutralization involves the reaction between an acid and a base, producing water and a salt. In buffer solutions, the weak acid and its conjugate base are poised to neutralize any added strong acid or base, respectively. For example, when a small amount of hydrochloric acid (HCl) is added to a buffer, the conjugate base (A^-) present in the buffer will neutralize HCl:

\[ A^- + HCl \rightarrow HA + Cl^- \]

Similarly, if a strong base like sodium hydroxide (NaOH) is introduced, the weak acid component (HA) of the buffer comes into play:

\[ HA + NaOH \rightarrow H_2O + Na^+ + A^- \]

It's this delicate dance of weak acid and base neutralizing their stronger counterparts that prevents significant swings in the pH of a buffer solution. For students, it's vital to grasp that while the buffer doesn't prevent the acid or base from being added, it essentially 'absorbs' the pH impact through this neutralization process.
Buffering Capacity
Buffering capacity refers to the amount of acid or base that a buffer solution can neutralize before its pH begins to change significantly. It is the buffer's quantitative aspect of resistance to pH changes. High buffering capacity is indicative of a solution's ability to withstand substantial intrusions of external acids or bases without surrendering its pH stability. This capacity is influenced by two major factors: the absolute concentrations of the buffering agents—both the weak acid and conjugate base—and their relative ratio.

As the concentration of these agents increases, the buffering capacity of the solution goes up. This is because there are more molecules of the weak acid and conjugate base available to engage in neutralization reactions. As students often seek simple takeaways, remember, high concentrations of buffer components equal high buffering capacity.

However, maintain caution with concentration; while a good buffer has high concentrations, these need to be managed alongside maintaining the equilibrium between the weak acid and conjugate base, following the Henderson-Hasselbalch equation. A balance between 'quantity' and 'proportional equilibrium' is the secret sauce to a buffer solution that can robustly resist pH alterations.

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Most popular questions from this chapter

Repeat the procedure in Exercise 61 , but for the titration of \(25.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NH}_{3}\left(K_{\mathrm{b}}=1.8 \times 10^{-5}\right)\) with \(0.100 \mathrm{M}\) \(\mathrm{HCl}\)

Calculate the \(\mathrm{pH}\) of a solution that is \(0.60 \mathrm{M} \mathrm{HF}\) and \(1.00 M \mathrm{KF}\).

A student titrates an unknown weak acid, HA, to a pale pink phenolphthalein end point with \(25.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\). The student then adds \(13.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}\). The \(\mathrm{pH}\) of the resulting solution is \(4.70\). How is the value of \(\mathrm{p} K_{\mathrm{a}}\) for the unknown acid related to 4.70?

a. Calculate the \(\mathrm{pH}\) of a buffered solution that is \(0.100 \mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\) (benzoic acid, \(K_{\mathrm{a}}=6.4 \times 10^{-5}\) ) and \(0.100 \mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{Na}\). b. Calculate the \(\mathrm{pH}\) after \(20.0 \%\) (by moles) of the benzoic acid is converted to benzoate anion by addition of a strong base. Use the dissociation equilibrium $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}^{+}(a q) $$ to calculate the \(\mathrm{pH}\). c. Do the same as in part b, but use the following equilibrium to calculate the \(\mathrm{pH}\) : \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{OH}^{-}(a q)\) d. Do your answers in parts \(\mathrm{b}\) and \(\mathrm{c}\) agree? Explain.

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.)

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