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Chapter 17: Problem 107

The effect of adding \(0.001 \mathrm{mol} \mathrm{KOH}\) to 1.00 Lof a solution that is \(0.10 \mathrm{M} \mathrm{NH}_{3}-0.10 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) is to (a) raise the pH very slightly; (b) lower the pH very slightly; (c) raise the pH by several units; (d) lower the pH by several units.

Short Answer

Expert verified
The effect of adding \(0.001 \mathrm{mol} \mathrm{KOH}\) to 1.00 L of a solution that is \(0.10 \mathrm{M} \mathrm{NH}_{3}-0.10 \mathrm{M} \mathrm{NH}_{4}\mathrm{Cl}\) is to raise the pH very slightly.

Step by step solution

01

Determine reaction of buffer solution

When \(0.001 \mathrm{mol} \mathrm{KOH}\) (a strong base) is added to the buffer solution, it reacts with the weak acid (ammonium ion) in the buffer to form more ammonia and water. This is shown in the below equation \[ \mathrm{NH}_{4}^{+} + OH^{-} -> \mathrm{NH}_{3} + H_{2}O \]
02

Calculate new concentrations of buffer components

Calculate the new concentrations of buffer components. The concentration of \( \mathrm{NH}_{4}^{+}\) ions would decrease by number of moles of \(KOH\) = \(0.10 - 0.001 = 0.099 M\). The concentration of \( \mathrm{NH}_{3}\) would increase by number of moles of \(KOH\) = \(0.10 + 0.001 = 0.101 M\).
03

Use the Henderson-Hasselbalch equation to calculate change in pH

Now, use the Henderson-Hasselbalch equation for buffers, \(pH = pK_a + log10([base]/[acid])\). The \(pK_a\) value of \( \mathrm{NH}_{4}^{+}\) is 9.25. Substitute the new concentrations into the equation: \(pH = 9.25 + log10(0.101/0.099) = 9.25 + 0.00899 = 9.259. Thus, the pH slightly increases after addition of KOH.

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Key Concepts

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

Henderson-Hasselbalch Equation
The Henderson-Hasselbalch Equation is a fundamental tool for anyone learning about buffer solutions in chemistry. It's used to estimate the pH of a buffer solution containing a weak acid and its conjugate base. Here's the equation: \[pH = pK_a + \log\left(\frac{[\text{base}]}{[\text{acid}]}\right)\]Where:

  • \(pH\) is the measure of acidity or basicity of the solution.

  • \(pK_a\) is the negative log of the acid dissociation constant, a measure of the strength of the weak acid.

  • \([\text{base}]\) is the concentration of the base form of the buffer.

  • \([\text{acid}]\) is the concentration of the acid form of the buffer.

This equation is derived from the equilibrium expression for the dissociation of a weak acid, allowing us to relate pH changes to the ratio of the concentrations of the base and acid components in a buffer. It highlights how buffers resist pH changes by adjusting the acid and base ratio as acids or bases are added.
pH Calculation
Calculating pH in buffer solutions can seem tricky, but once you understand the process, it becomes straightforward. In a buffer solution, acids and bases are not completely dissociated, allowing for a stable pH. To calculate pH using the Henderson-Hasselbalch equation, you need the pKa of the weak acid and the concentrations of the acid and base molecules.

Take for instance ammonia (\(\mathrm{NH}_3\)), which forms a buffer with its conjugate acid, ammonium (\(\mathrm{NH}_4^+\)). If the concentration of \(\mathrm{NH}_3\) slightly increases and \(\mathrm{NH}_4^+\) slightly decreases due to added KOH, you can plug these values into the equation to find the new pH.

For example, given:

  • The concentration of \(\mathrm{NH}_3\) is 0.101 M.

  • The concentration of \(\mathrm{NH}_4^+\) is 0.099 M.

Substitute into:\[ pH = 9.25 + \log\left(\frac{0.101}{0.099}\right) \]Simplifying gives:\[ pH = 9.25 + 0.009 \]\[ pH = 9.259 \]This shows a slight increase in pH, demonstrating the buffer's ability to dampen drastic pH changes.
Acid-Base Equilibrium
Understanding acid-base equilibrium is crucial for working with buffer solutions. In any buffer solution, there exists an equilibrium between the weak acid and its conjugate base. This dynamic balance allows buffers to resist drastic pH changes when small amounts of strong acids or bases are added.
This is because:
  • Adding a base shifts the equilibrium towards forming more of the acid form, reducing pH increase.
  • Adding an acid shifts the equilibrium towards forming more of the base form, reducing pH decrease.
The equilibrium can be represented by the reversible equation:\[ \mathrm{NH}_{4}^{+} + \mathrm{OH}^{-} \rightleftharpoons \mathrm{NH}_3 + \mathrm{H}_2\mathrm{O} \]When you introduce KOH into the buffer solution, the hydroxide ions \(\mathrm{OH}^-\) react with ammonium ions \(\mathrm{NH}_4^+\), converting them to ammonia \(\mathrm{NH}_3\).

This highlights how buffers maintain equilibrium by adjusting ion concentrations and hence stabilizing the pH, ensuring that any change remains minimal and within a predictable range.

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Most popular questions from this chapter

Amino acids contain both an acidic carboxylic acid group \((-\mathrm{COOH})\) and a basic amino group \(\left(-\mathrm{NH}_{2}\right)\) The amino group can be protonated (that is, it has an extra proton attached) in a strongly acidic solution. This produces a diprotic acid of the form \(\mathrm{H}_{2} \mathrm{A}^{+}\), as exemplified by the protonated amino acid alanine. The protonated amino acid has two ionizable protons that can be titrated with \(\mathrm{OH}^{-}\) For the \(-\mathrm{COOH}\) group, \(\mathrm{pK}_{\mathrm{a}_{1}}=2.34 ;\) for the \(-\mathrm{NH}_{3}^{+}\) group, \(\mathrm{p} K_{\mathrm{a}_{2}}=9.69 .\) Consider the titration of a 0.500 M solution of alanine hydrochloride with \(0.500 \mathrm{M} \mathrm{NaOH}\) solution. What is the \(\mathrm{pH}\) of \((\mathrm{a})\) the \(0.500 \mathrm{M}\) alanine hydrochloride; (b) the solution at the first half- neutralization point; (c) the solution at the first equivalence point? The dominant form of alanine present at the first equivalence point is electrically neutral despite the positive charge and negative charge it possesses. The point at which the neutral form is produced is called the isoelectric point. Confirm that the \(\mathrm{pH}\) at the isoelectric point is \(\mathrm{pH}=\frac{1}{2}\left(\mathrm{pK}_{\mathrm{a}_{1}}+\mathrm{p} \mathrm{K}_{\mathrm{a}_{2}}\right)\) What is the \(\mathrm{pH}\) of the solution (d) halfway between the first and second equivalence points? (e) at the second equivalence point? (f) Calculate the pH values of the solutions when the following volumes of the \(0.500 \mathrm{M} \mathrm{NaOH}\) have been added to \(50 \mathrm{mL}\) of the \(0.500 \mathrm{M}\) alanine hydrochloride solution: \(10.0 \mathrm{mL}, 20.0 \mathrm{mL}, 30.0 \mathrm{mL}, 40.0 \mathrm{mL}, 50.0 \mathrm{mL}\) \(60.0 \mathrm{mL}, 70.0 \mathrm{mL}, 80.0 \mathrm{mL}, 90.0 \mathrm{mL}, 100.0 \mathrm{mL},\) and \(110.0 \mathrm{mL}\) (g) Sketch the titration curve for the 0.500 M solution of alanine hydrochloride, and label significant points on the curve.

Solution (a) is \(100.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}\) and solution (b) is \(150.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{NaCH}_{3} \mathrm{COO}\). A few drops of thymol blue indicator are added to each solution. What is the color of each solution? What is the color of the solution obtained when these two solutions are mixed?

In 1922 Donald D. van Slyke ( J. Biol. Chem., 52, 525) defined a quantity known as the buffer index: \(\beta=\mathrm{d} C_{\mathrm{b}} / \mathrm{d}(\mathrm{pH}),\) where \(\mathrm{d} C_{\mathrm{b}}\) represents the increment of moles of strong base to one liter of the buffer. For the addition of a strong acid, he wrote \(\beta=-\mathrm{d} C_{\mathrm{a}} / \mathrm{d}(\mathrm{pH})\) By applying this idea to a monoprotic acid and its conjugate base, we can derive the following expression: \(\beta=2.303\left(\frac{K_{w}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]+\frac{\mathrm{CK}_{\mathrm{a}}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left(\mathrm{K}_{\mathrm{a}}+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\right)^{2}}\right)\) where \(C\) is the total concentration of monoprotic acid and conjugate base. (a) Use the above expression to calculate the buffer index for the acetic acid buffer with a total acetic acid and acetate ion concentration of \(2.0 \times 10^{-2}\) and a \(\mathrm{pH}=5.0\) (b) Use the buffer index from part (a) and calculate the \(\mathrm{pH}\) of the buffer after the addition of of a strong acid. (Hint: Let \(\left.\mathrm{d} C_{\mathrm{a}} / \mathrm{d}(\mathrm{pH}) \approx \Delta C_{\mathrm{a}} / \Delta \mathrm{pH} .\right)\) (c) Make a plot of \(\beta\) versus \(\mathrm{pH}\) for a \(0.1 \mathrm{M}\) acetic acid buffer system. Locate the maximum buffer index as well as the minimum buffer indices.

To increase the ionization of formic acid, \(\mathrm{HCOOH}(\mathrm{aq})\) which of the following should be added to the solution? (a) \(\mathrm{NaCl} ;\) (b) \(\mathrm{NaHCOO} ;(\mathrm{c}) \mathrm{H}_{2} \mathrm{SO}_{4} ;\) (d) \(\mathrm{NaHCO}_{3}\)

Determine the following characteristics of the titration curve for \(20.0 \mathrm{mL}\) of \(0.275 \mathrm{M} \mathrm{NH}_{3}(\mathrm{aq})\) titrated with \(0.325 \mathrm{M} \mathrm{HI}(\mathrm{aq})\) (a) the initial \(\mathrm{pH}\) (b) the volume of \(0.325 \mathrm{M} \mathrm{HI}(\mathrm{aq})\) at the equivalence point (c) the \(\mathrm{pH}\) at the half-neutralization point (d) the \(\mathrm{pH}\) at the equivalence point
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