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Chapter 15: Problem 12

Consider a buffer solution where [weak acid] \(>\) [conjugate base]. How is the pH of the solution related to the \(\mathrm{p} K_{\mathrm{a}}\) value of the weak acid? If [conjugate base \(]>[\text { weak acid }]\) , how is pH related to \(\mathrm{p} K_{\mathrm{a}} ?\)

Short Answer

Expert verified
In a buffer solution, the pH is related to the pKa value of the weak acid following the Henderson-Hasselbalch equation: \( pH = pKa + \log \left(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\right) \). If [weak acid] > [conjugate base], the fraction inside the log is less than 1, resulting in a negative log value and a pH lower than the pKa. Conversely, if [conjugate base] > [weak acid], the fraction inside the log is greater than 1, leading to a positive log value and a pH higher than the pKa.

Step by step solution

01

Understand the Buffer Solution

A buffer solution is an aqueous solution that consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. The purpose of a buffer solution is to maintain a relatively constant pH when a small amount of acid or base is added to it.
02

Recall the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a way to determine the pH of a buffer solution. The equation is given by: \[ pH = pKa + \log \left(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\right) \]
03

Find the relationship between pH and pKa for [weak acid] > [conjugate base]

Using the Henderson-Hasselbalch equation: \[ pH = pKa + \log \left(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\right) \] Since [weak acid] > [conjugate base], the fraction inside the log will be less than 1. The log of a number smaller than 1 is negative, so the pH will be lower than the pKa in this scenario.
04

Find the relationship between pH and pKa for [conjugate base] > [weak acid]

Again, using the Henderson-Hasselbalch equation: \[ pH = pKa + \log \left(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\right) \] In this case, since [conjugate base] > [weak acid], the fraction inside the log will be greater than 1. The log of a number greater than 1 is positive, so the pH will be higher than the pKa in this situation. In conclusion, if [weak acid] > [conjugate base], the pH of the buffer solution will be lower than the pKa value of the weak acid, while if [conjugate base] > [weak acid], the pH of the buffer solution will be higher than the pKa value of the weak acid.

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Most popular questions from this chapter

Sketch a pH curve for the titration of a weak acid (HA) with a strong base (NaOH). List the major species, and explain how you would go about calculating the pH of the solution at various points, including the halfway point and the equivalence point

You have 75.0 \(\mathrm{mL}\) of 0.10 \(\mathrm{M}\) HA. After adding 30.0 \(\mathrm{mL}\) of 0.10 \(\mathrm{M} \mathrm{NaOH}\) , the \(\mathrm{pH}\) is \(5.50 .\) What is the \(K_{\mathrm{a}}\) value of \(\mathrm{HA}\) ?

Calculate the pH after 0.010 mole of gaseous HCl is added to 250.0 \(\mathrm{mL}\) of each of the following buffered solutions. $$ \begin{array}{l}{\text { a. } 0.050 M \mathrm{NH}_{3} / 0.15 \mathrm{MNH}_{4} \mathrm{Cl}} \\ {\text { b. } 0.50 \mathrm{M} \mathrm{NH}_{3} / 1.50 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}}\end{array} $$ Do the two original buffered solutions differ in their pH or their capacity? What advantage is there in having a buffer with a greater capacity?

Consider the titration of 40.0 \(\mathrm{mL}\) of 0.200 $\mathrm{M} \mathrm{HClO}_{4}$ by 0.100 \(\mathrm{M}\) KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 80.0 \mathrm{mL}} \\ {\text { b. } 10.0 \mathrm{mL}} & {\text { e. } 100.0 \mathrm{mL}} \\ {\text { c. } 40.0 \mathrm{mL}}\end{array} $$

a. Calculate the \(\mathrm{pH}\) of a buffered solution that is 0.100 \(\mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\) (benzoic acid, \(K_{\mathrm{a}}=6.4 \times 10^{-5} )\) and 0.100 \(\mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{Na}\) b. Calculate the pH after 20.0\(\%\) (by moles) of the benzoic acid is converted to benzoate anion by addition of a strong base. Use the dissociation equilibrium $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}^{+}(a q) $$ to calculate the pH. c. Do the same as in part b, but use the following equilibrium to calculate the \(\mathrm{pH} :\) $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{OH}^{-}(a q) $$ d. Do your answers in parts \(b\) and \(c\) agree? Explain.
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