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Chapter 16: Problem 69

Which of the following solutions has the greater buffer capacity: (a) \(50 \mathrm{~mL}\) of \(0.20 \mathrm{M} \mathrm{NH}_{4} \mathrm{Br}-0.30 \mathrm{M} \mathrm{NH}_{3}\) or (b) \(50 \mathrm{~mL}\) of \(0.40 \mathrm{M} \mathrm{NH}_{4} \mathrm{Br}-0.60 \mathrm{M} \mathrm{NH}_{3} ?\) Explain.

Short Answer

Expert verified
Solution (b) has the greater buffer capacity because it has higher component concentrations.

Step by step solution

01

Understand What Buffer Capacity Means

Buffer capacity is a measure of a buffer solution's ability to resist pH changes when acids or bases are added. The greater the concentration of the buffer components, the greater the buffer capacity.
02

Analyze the Given Buffer Solutions

We are provided with two buffer solutions. Solution (a) contains \(0.20 \text{ M NH}_4\text{Br}\) and \(0.30 \text{ M NH}_3\). Solution (b) consists of \(0.40 \text{ M NH}_4\text{Br}\) and \(0.60 \text{ M NH}_3\).
03

Compare Total Molar Concentrations

To determine which solution has a greater buffer capacity, compare the total molar concentrations of the acidic and basic components of both solutions. For solution (a), the concentration of \(\text{NH}_4^+\) is 0.20 M and \(\text{NH}_3\) is 0.30 M, giving a total of \(0.20 + 0.30 = 0.50 \text{ M}\). For solution (b), we have 0.40 M \(\text{NH}_4^+\) and 0.60 M \(\text{NH}_3\), hence a total of \(0.40 + 0.60 = 1.00 \text{ M}\).
04

Identify the Solution with Higher Capacity

Since buffer capacity increases with the increase in the total concentration of buffer components, and solution (b) has a higher total concentration (1.00 M) than solution (a) (0.50 M), hence solution (b) has the greater buffer capacity.

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Key Concepts

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

Buffer Solutions
A buffer solution is specially designed to maintain a stable pH even when small amounts of an acid or base are added. This is crucial in many biological and chemical processes where consistent pH levels are necessary. Buffer solutions are usually made up of a weak acid and its conjugate base, or a weak base and its conjugate acid. In our problem, the buffer comprises ammonia (\( ext{NH}_3\)) and ammonium bromide (\( ext{NH}_4 ext{Br}\)).

These components work together to neutralize added acids or bases. For instance, if a small amount of acid is added, the weak base part of the buffer will neutralize it, and vice versa. Thus, the solution's pH remains relatively constant. The concept of buffer capacity is vital here because it quantitatively describes how much acid or base the buffer can neutralize before significant pH changes occur.
Molar Concentration
Molar concentration refers to the amount of a solute, measured in moles, dissolved in a specific volume of solution, measured in liters. It is expressed in \( ext{M} \), or molarity, which stands for moles per liter. Understanding molarity is essential to understanding buffer capacity.

In the provided exercise, we compare two solutions with different molar concentrations of buffer components. Solution (a) has molarities of \(0.20 ext{ M}\) for \( ext{NH}_4 ext{Br}\) and \(0.30 ext{ M}\) for \( ext{NH}_3\). On the other hand, solution (b) has molarities of \(0.40 ext{ M}\) for \( ext{NH}_4 ext{Br}\) and \(0.60 ext{ M}\) for \( ext{NH}_3\).

The solution with the higher total molar concentration has greater capacity to resist pH changes because it contains more buffering agents to counteract the effects of added acids or bases.
Acid-Base Equilibrium
Acid-base equilibrium involves the balance between acids and bases in a solution. This equilibrium is crucial in buffer solutions to ensure the mixture can effectively resist changes in pH. The key components that maintain this balance are the weak acids and their conjugate bases or weak bases and their conjugate acids.

For instance, in the solutions discussed, \( ext{NH}_4^+\) acts as the conjugate acid while \( ext{NH}_3\) acts as the weak base. When an acid is introduced, the equilibrium allows the \( ext{NH}_3\) to react with the \( ext{H}^+\) ions, forming more \( ext{NH}_4^+\). Conversely, when a base is added, the \( ext{NH}_4^+\) can donate a proton to neutralize it.

Understanding this equilibrium helps explain why solution (b) has a greater buffer capacity鈥攊ts higher concentrations allow it to neutralize more added acid or base, maintaining the desired pH balance for longer periods.

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Most popular questions from this chapter

You need a buffer solution that has \(\mathrm{pH}=7.00\). Which of the following buffer systems should you choose? Explain. (a) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) (b) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) (c) \(\mathrm{HPO}_{4}^{2-}\) and \(\mathrm{PO}_{4}^{3-}\)

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Consider the titration of \(25.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HCO}_{2} \mathrm{H}\) with \(0.250\) M NaOH. How many milliliters of base are required to reach the equivalence point? Calculate the \(\mathrm{pH}\) at each of the following points: (a) After the addition of \(7.0 \mathrm{~mL}\) of base (b) Halfway to the equivalence point (c) At the equivalence point (d) After the addition of \(25.0 \mathrm{~mL}\) of base

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\(\mathrm{Ca}^{2+}\), which causes clotting, is removed from donated blood by precipitation with sodium oxalate \(\left(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right) . \mathrm{CaC}_{2} \mathrm{O}_{4}\) is a sparingly soluble salt \(\left(K_{\mathrm{sp}}=2.3 \times 10^{-9}\right) .\) If the desired \(\left[\mathrm{Ca}^{2+}\right]\) is less than \(3.0 \times 10^{-8} \mathrm{M}\), what must be the minimum concentration of \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) in the blood sample?
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