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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 NH鈧 and NH鈧凜l is not possible because the calculated ratio of [NH鈧僝/[NH鈧勨伜] 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鈦籡/[HA]) where pH is the desired pH of the solution, pKa is the negative logarithm of the acid ionization constant (Ka), [A鈦籡 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鈦籡 and [HA] with the concentrations of NH鈧 and NH鈧勨伜, respectively.
02

Calculate the pKa of NH鈧勨伜

The ionization constant, Kb, of NH鈧 is 1.8 脳 10鈦烩伒. To find the pKa of the conjugate acid, NH鈧勨伜, 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 脳 10鈦宦光伌. To calculate the pKa, we will take the negative logarithm of the Ka.
03

Solve the Henderson-Hasselbalch equation for [A鈦籡/[HA]

Substitute the values of pH and pKa in the equation: 6.50 = pKa + log ([NH鈧僝/[NH鈧勨伜]) Solve the equation to obtain the ratio [NH鈧僝/[NH鈧勨伜].
04

Determine if it's possible to obtain a pH of 6.50

If the calculated ratio [NH鈧僝/[NH鈧勨伜] is feasible through mixing NH鈧 and NH鈧凜l, 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鈦).
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鈦) 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.
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鈦), 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鈦. Conversely, when A鈦 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鈦) 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.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}\) and \(1.00\) atm pressure, one liter of ammonia is bubbled into \(725 \mathrm{~mL}\) of water. Assume that all the ammonia dissolves and the volume of the solution is the volume of the water. A \(50.0-\mathrm{mL}\) portion of the prepared solution is titrated with \(0.2193 \mathrm{M} \mathrm{HNO}_{3} .\) Calculate the \(\mathrm{pH}\) of the solution (a) before titration. (b) halfway to the equivalence point. (c) at the equivalence point.

Write a balanced net ionic equation for the reaction of each of the following aqueous solutions with \(\mathrm{H}^{+}\) ions. (a) sodium fluoride (b) barium hydroxide (c) potassium dihydrogen phosphate \(\left(\mathrm{KH}_{2} \mathrm{PO}_{4}\right)\)

For an aqueous solution of acetic acid to be called "distilled white vinegar" it must contain \(5.0 \%\) acetic acid by mass. A solution with a density of \(1.05 \mathrm{~g} / \mathrm{mL}\) has a \(\mathrm{pH}\) of \(2.95 .\) Can the solution be called "distilled white vinegar"?

Explain why (a) the pH decreases when lactic acid is added to a sodium lactate solution. (b) the \(\mathrm{pH}\) of \(0.1 \mathrm{M} \mathrm{NH}_{3}\) is less than \(13.0\). (c) a buffer resists changes in pH caused by the addition of \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-} .\) (d) a solution with a low \(\mathrm{pH}\) is not necessarily a strong acid solution.

A \(0.4000 \mathrm{M}\) solution of nitric acid is used to titrate \(50.00 \mathrm{~mL}\) of \(0.237 \mathrm{M}\) barium hydroxide. (Assume that volumes are additive.) (a) Write a balanced net ionic equation for the reaction that takes place during titration. (b) What are the species present at the equivalence point? (c) What volume of nitric acid is required to reach the equivalence point? (d) What is the \(\mathrm{pH}\) of the solution before any \(\mathrm{HNO}_{3}\) is added? (e) What is the \(\mathrm{pH}\) of the solution halfway to the equivalence point? (f) What is the \(\mathrm{pH}\) of the solution at the equivalence point?

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