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Chapter 14: Problem 10

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] \(>\) [weak acid], how is pH related to \(\mathrm{p} K_{\mathrm{a}} ?\)

Short Answer

Expert verified
In a buffer solution, the relationship between pH and pKa can be described using the Henderson-Hasselbalch equation: \(pH = pKa + \log\frac{[\text{conjugate base}]}{[\text{weak acid}]}\). When [weak acid] > [conjugate base], the pH of the solution will be lower than the pKa of the weak acid. Conversely, when [conjugate base] > [weak acid], the pH of the solution will be higher than the pKa of the weak acid.

Step by step solution

01

Henderson-Hasselbalch equation

This equation is given as follows: \(pH = pKa + \log\frac{[\text{conjugate base}]}{[\text{weak acid}]}\) Now, let's analyze the given scenarios. Case 1: The concentration of the weak acid is greater than the concentration of the conjugate base.
02

Analyzing the relationship

In this case, since [weak acid] > [conjugate base], the fraction \(\frac{ [\text{conjugate base}]}{[\text{weak acid}]}\) will be less than 1. Consequently, the value of the logarithm will be negative because \(\log(x) < 0\) if \(0 < x < 1\). Therefore, in this scenario, the pH of the solution will be lower than the pKa of the weak acid. Case 2: The concentration of the conjugate base is greater than the concentration of the weak acid.
03

Analyzing the relationship

In this case, since [conjugate base] > [weak acid], the fraction \(\frac{ [\text{conjugate base}]}{[\text{weak acid}]}\) will be greater than 1. Consequently, the value of the logarithm will be positive because \(\log(x) > 0\) if \(x > 1\). Therefore, in this scenario, the pH of the solution will be higher than the pKa of the weak acid. In conclusion, when [weak acid] > [conjugate base], the pH of the solution will be lower than the pKa of the weak acid. And when [conjugate base] > [weak acid], the pH of the solution will be higher than the pKa of the weak acid.

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Most popular questions from this chapter

You have a solution of the weak acid HA and add some of the salt NaA to it. What are the major species in the solution? What do you need to know to calculate the \(\mathrm{pH}\) of the solution, and how would you use this information? How does the pH of the solution of just the HA compare with that of the final mixture? Explain.

Consider the titration of \(40.0 \mathrm{mL}\) of \(0.200 \mathrm{M} \mathrm{HClO}_{4}\) by \(0.100 M\) KOH. Calculate the \(p\) H of the resulting solution after the following volumes of KOH have been added. a. \(0.0 \mathrm{mL}\) b. \(10.0 \mathrm{mL}\) c. \(40.0 \mathrm{mL}\) d. \(80.0 \mathrm{mL}\) e. \(100.0 \mathrm{mL}\)

Repeat the procedure in Exercise \(61,\) but for the titration of \(25.0 \mathrm{mL}\) of 0.100 \(\mathrm{M}\) \(\mathrm{HNO}_{3}\) with 0.100 \(\mathrm{M}\) \(\mathrm{NaOH}\).

Consider the following four titrations (i-iv): i. \(150 \mathrm{mL}\) of \(0.2 \mathrm{M} \mathrm{NH}_{3}\left(K_{\mathrm{b}}=1.8 \times 10^{-5}\right)\) by \(0.2 \mathrm{M} \mathrm{HCl}\) ii. \(150 \mathrm{mL}\) of \(0.2 \mathrm{M}\) HCl by \(0.2 \mathrm{M} \mathrm{NaOH}\) iii. \(150 \mathrm{mL}\) of \(0.2 \mathrm{M} \mathrm{HOCl}\left(K_{\mathrm{a}}=3.5 \times 10^{-8}\right)\) by \(0.2 \mathrm{M} \mathrm{NaOH}\) iv. \(150 \mathrm{mL}\) of \(0.2 \mathrm{M} \mathrm{HF}\left(K_{\mathrm{a}}=7.2 \times 10^{-4}\right)\) by \(0.2 \mathrm{M} \mathrm{NaOH}\) a. Rank the four titrations in order of increasing \(\mathrm{pH}\) at the halfway point to equivalence (lowest to highest \(\mathrm{pH}\) ). b. Rank the four titrations in order of increasing \(\mathrm{pH}\) at the equivalence point. c. Which titration requires the largest volume of titrant (HCl or \(\mathrm{NaOH}\) ) to reach the equivalence point?

Calculate the \(\mathrm{pH}\) of each of the following solutions. a. \(0.100 M\) HONH \(_{2}\left(K_{\mathrm{b}}=1.1 \times 10^{-8}\right)\) b. \(0.100 M\) HONH \(_{3}\) Cl c. pure \(\mathrm{H}_{2} \mathrm{O}\) d. a mixture containing 0.100 \(M \mathrm{HONH}_{2}\) and \(0.100 \mathrm{M}\) \(\mathrm{HONH}_{3} \mathrm{Cl}\)
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