Chapter 14: Problem 73
Explain why it is not possible to prepare a buffer with a pH of \(6.50\) by mixing \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4} \mathrm{Cl}\).
Short Answer
Expert verified
Answer: The preparation of a buffer with a pH of 6.50 using NH₃ and NH₄Cl is not possible because the calculated ratio of [NH₃]/[NH₄⁺] is not feasible, making it impossible to achieve the desired pH using these components.
Step by step solution
01
Write the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A⁻]/[HA])
where pH is the desired pH of the solution, pKa is the negative logarithm of the acid ionization constant (Ka), [A⁻] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
In this case, we will replace the pH with 6.50 and the terms [A⁻] and [HA] with the concentrations of NH₃ and NH₄⁺, respectively.
02
Calculate the pKa of NH₄⁺
The ionization constant, Kb, of NH₃ is 1.8 × 10⁻⁵. To find the pKa of the conjugate acid, NH₄⁺, we will first calculate Ka using the relationship between Ka and Kb:
Ka * Kb = Kw
where Kw is the ionization constant of water, which is 1.0 × 10⁻¹⁴.
To calculate the pKa, we will take the negative logarithm of the Ka.
03
Solve the Henderson-Hasselbalch equation for [A⁻]/[HA]
Substitute the values of pH and pKa in the equation:
6.50 = pKa + log ([NH₃]/[NH₄⁺])
Solve the equation to obtain the ratio [NH₃]/[NH₄⁺].
04
Determine if it's possible to obtain a pH of 6.50
If the calculated ratio [NH₃]/[NH₄⁺] is feasible through mixing NH₃ and NH₄Cl, it implies that it's possible to obtain a buffer with a pH of 6.50. If not, it would be impossible to prepare a buffer with the given pH using these chemicals.
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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 pivotal in biochemistry and chemistry, especially when dealing with buffer solutions. The equation is a rearranged form of the acid dissociation constant formula and it's expressed as:
\[\text{pH} = \text{pKa} + \log\left(\frac{[A^-]}{[HA]}\right)\]
In this notation, pH represents the acidity level of the solution, pKa is the negative logarithm of the acid ionization constant (Ka) which provides a measure of the acid's strength, [A^-] denotes the concentration of the conjugate base, and [HA] stands for the concentration of the weak acid. The equation is particularly useful when you need to find the pH of a buffer solution.
One common misinterpretation is assuming that any mixture of a weak acid and its conjugate base can achieve a wide range of pH levels. However, the pH achieved heavily depends on the pKa and the ratio of the conjugate base to acid concentrations. In case of ammonia (NH3) and ammonium chloride (NH4Cl), which pair up as a conjugate acid-base in water, the resulting pH can only span a limited range that centers around the pKa value of the ammonium ion (NH4⁺).
\[\text{pH} = \text{pKa} + \log\left(\frac{[A^-]}{[HA]}\right)\]
In this notation, pH represents the acidity level of the solution, pKa is the negative logarithm of the acid ionization constant (Ka) which provides a measure of the acid's strength, [A^-] denotes the concentration of the conjugate base, and [HA] stands for the concentration of the weak acid. The equation is particularly useful when you need to find the pH of a buffer solution.
One common misinterpretation is assuming that any mixture of a weak acid and its conjugate base can achieve a wide range of pH levels. However, the pH achieved heavily depends on the pKa and the ratio of the conjugate base to acid concentrations. In case of ammonia (NH3) and ammonium chloride (NH4Cl), which pair up as a conjugate acid-base in water, the resulting pH can only span a limited range that centers around the pKa value of the ammonium ion (NH4⁺).
Acid-Base Equilibrium
The concept of acid-base equilibrium is essential in understanding buffer systems. In an aqueous solution, an acid-base equilibrium is established when a weak acid donates protons (H⁺) to water, forming its conjugate base, and a weak base accepts protons from water, forming its conjugate acid. This reversible process is captured by the equilibrium expression for an acid
\[\text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{A}^- + \text{H}_3\text{O}^+\]
and for a base
\[\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{HB}^+ + \text{OH}^-\].
Each side of the equilibrium has to balance out in terms of charge and mass. The equilibrium constant (Ka for the acids and Kb for the bases) quantifies the extent to which the acid or base ionizes in solution. For a buffer system, it's the ratio of the concentrations of the conjugate acid-base pair that determines the pH, rather than the absolute concentrations of the acid and base. Therefore, manipulating this ratio can alter the pH, but only within the limits set by the acid dissociation constant of the weak acid involved in the buffer system.
\[\text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{A}^- + \text{H}_3\text{O}^+\]
and for a base
\[\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{HB}^+ + \text{OH}^-\].
Each side of the equilibrium has to balance out in terms of charge and mass. The equilibrium constant (Ka for the acids and Kb for the bases) quantifies the extent to which the acid or base ionizes in solution. For a buffer system, it's the ratio of the concentrations of the conjugate acid-base pair that determines the pH, rather than the absolute concentrations of the acid and base. Therefore, manipulating this ratio can alter the pH, but only within the limits set by the acid dissociation constant of the weak acid involved in the buffer system.
Conjugate Acid-Base Pairs
Conjugate acid-base pairs are at the heart of buffer solutions and their capability to resist changes in pH. Each conjugate pair consists of a weak acid (HA) and its corresponding weak base (A⁻), connected by the gain or loss of a proton. This relation is illustrated in the ionization of the acid:
\[\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\].
In this process, when HA donates a proton, it becomes its conjugate base A⁻. Conversely, when A⁻ accepts a proton, it forms its conjugate acid HA. This dynamic is essential in buffer solutions because it allows the absorption or donation of protons, as needed, to maintain a relatively constant pH.
However, it's important to note that the pKa of the weak acid in the pair dictates the pH range that the buffer can effectively regulate. To reiterate, the acid-base pair of ammonia and ammonium (NH3/NH4⁺) can only serve as an effective buffer in a pH range that's near the pKa of ammonium. If the desired pH is significantly different from the pKa, the buffering capacity is lost, and achieving the target pH becomes impractical with that specific conjugate acid-base pair.
\[\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\].
In this process, when HA donates a proton, it becomes its conjugate base A⁻. Conversely, when A⁻ accepts a proton, it forms its conjugate acid HA. This dynamic is essential in buffer solutions because it allows the absorption or donation of protons, as needed, to maintain a relatively constant pH.
However, it's important to note that the pKa of the weak acid in the pair dictates the pH range that the buffer can effectively regulate. To reiterate, the acid-base pair of ammonia and ammonium (NH3/NH4⁺) can only serve as an effective buffer in a pH range that's near the pKa of ammonium. If the desired pH is significantly different from the pKa, the buffering capacity is lost, and achieving the target pH becomes impractical with that specific conjugate acid-base pair.