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Chapter 1: Problem 21

A laboratory technician wishes to create a buffered solution with a pH of 5. Which of the following acids would be the best choice for the buffer? (A) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \quad K_{a}=5.9 \times 10^{-2}\) (B) \(\mathrm{H}_{3} \mathrm{AsO}_{4} \quad K_{a}=5.6 \times 10^{-3}\) (C) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \quad K_{a}=1.8 \times 10^{-5}\) (D) \(\mathrm{HOCl}\) \(\quad K_{a}=3.0 \times 10^{-8}\)

Short Answer

Expert verified
Without the exact calculations, the specific answer cannot be provided. By applying the steps, the pKa of each acid can be determined. The acid that has the pKa value closest to the desired pH of 5 will be the best choice. This would be the short answer upon completing the calculations.

Step by step solution

01

Calculate the pKa of each acid

The pKa is calculated by applying the formula pKa = -log Ka. For each of the acids, calculate the pKa:\n (A) \( \mathrm{pKa} = - \log(5.9 \times 10^{-2}) \) \n (B) \(\mathrm{pKa} = - \log(5.6 \times 10^{-3})\) \n (C) \( \mathrm{pKa} = - \log(1.8 \times 10^{-5}) \) \n (D) \( \mathrm{pKa} = - \log(3.0 \times 10^{-8}) \)
02

Compare the resulting pKa values to the desired pH

After calculating the pKa values, compare each to the desired pH of 5. The acid with a pKa value closest to the desired pH is the best choice.
03

Identify the best acid for the buffer

By checking the calculated pKa values, the acid whose pKa is closest to 5 will be the most appropriate choice. If multiple acids have pKa values equally close to 5, any of them could be the best choice.

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Key Concepts

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

pKa Calculation
Understanding how to calculate the pKa of an acid is an essential skill in acid-base chemistry. The pKa value helps predict how an acid behaves in water, which is crucial for making buffer solutions. You calculate the pKa using the formula:
  • \[ \text{pKa} = -\log K_a \]
The given problem requires you to determine which acid is suitable for a buffer with a pH of 5. First, you calculate the pKa for each available acid using their respective dissociation constants \(K_a\):
  • (A) \[ \mathrm{pKa} = - \log(5.9 \times 10^{-2}) \]
  • (B) \[ \mathrm{pKa} = - \log(5.6 \times 10^{-3}) \]
  • (C) \[ \mathrm{pKa} = - \log(1.8 \times 10^{-5}) \]
  • (D) \[ \mathrm{pKa} = - \log(3.0 \times 10^{-8}) \]
By solving these, you get the pKa values for each acid, allowing you to find the one that's closest to the desired pH.
pH of Buffered Solutions
Buffer solutions are used to maintain a stable pH in a chemical environment. They are made from a weak acid and its conjugate base. For successful buffering, the acid's pKa should be close to the target pH.
In this case, to create a buffer with pH 5, you need to choose an acid whose pKa is near that value. This is because buffers are most effective when the pH is equal to the pKa of the acid. By calculating and comparing the pKa values, you determine which acid will best maintain this pH.
Once selected, the most suitable buffer will resist drastic pH changes, ensuring a stable environment. This is crucial in many laboratory and real-world applications where pH control is necessary.
Acid-Base Chemistry
In acid-base chemistry, understanding the behavior of acids and bases is foundational. The strength of an acid is determined by its ability to donate protons (H鈦� ions) in a solution, which is represented by its Ka value.
  • Ka, the acid dissociation constant: It indicates how completely an acid dissociates in water.
  • pKa, the negative log of Ka: A lower pKa means a stronger acid, as it dissociates more readily.
These concepts are key when working with buffer solutions. You need to match the pKa to the desired pH closely to make effective buffers.
Buffering involves equilibrium between the weak acid and its conjugate base, enabling it to neutralize added acids or bases. Mastering these principles allows chemists to control reactions and processes by keeping the pH within a narrow range.

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Most popular questions from this chapter

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ How many moles of electrons must be transferred to create 127 g of copper? (A) 1 mole of electrons (B) 2 moles of electrons (C) 3 moles of electrons (D) 4 moles of electrons

Questions 54-56 refer to the following. GRAPH CAN'T COPY Between propane and ethene, which will likely have the higher boiling point and why? (A) Propane, because it has a greater molar mass (B) Propane, because it has a more polarizable electron cloud (C) Ethene, because of the double bond (D) Ethene, because it is smaller in size

In which of the following reactions is entropy increasing? (A) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g)\) (B) \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)\) (C) \(\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{HCl}(g)\) (D) \(2 \mathrm{NO}_{2}(g) \rightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)\)

Which of the substances would be soluble in water? (A) Ethylene glycol only, because it has the longest bond lengths (B) Acetone only, because it is the most symmetrical (C) Ethanol and ethylene glycol only, because of their hydroxyl (-OH) (D) All three substances would be soluble in water due to their permanent dipoles.

\(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightarrow 2 \mathrm{HF}(g)\) Gaseous hydrogen and fluorine combine in the reaction above to form hydrogen fluride with an enthalpy change of \(-540 \mathrm{kJ}\) . What is the value of the heat of formation of \(\mathrm{HF}(g) ?\) (A) \(-1,080 \mathrm{kJ} / \mathrm{mol}\) (B) \(-270 \mathrm{kJ} / \mathrm{mol}\) (C) 270 \(\mathrm{kJ} / \mathrm{mol}\) (D) 540 \(\mathrm{kJ} / \mathrm{mol}\)
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