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Chapter 15: Problem 10

Consider a buffer solution where [weak acid] \(>\) [conjugate base]. How is the \(\mathrm{pH}\) 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 summary, when the concentration of the weak acid is greater than its conjugate base in a buffer solution, the pH of the solution is less than the pKa value. Conversely, when the concentration of the conjugate base is greater than the weak acid, the pH of the solution is greater than the pKa value. This relationship is derived from the Henderson-Hasselbalch equation: \[ pH = pKa + log \frac{[\text{conjugate base}]}{[\text{weak acid}]}\]

Step by step solution

01

Recall the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is used to determine the pH of a buffer solution: \[ pH = pKa + log \frac{[\text{conjugate base}]}{[\text{weak acid}]}\] where pH is the acidity of the solution, pKa refers to the combined dissociation constant of the weak acid, and the concentrations of the conjugate base and weak acid are given in moles per liter.
02

Determine the relationship between pH and pKa when [weak acid] > [conjugate base]

Given the condition [weak acid] > [conjugate base], we will use the Henderson-Hasselbalch equation: \[ pH = pKa + log \frac{[\text{conjugate base}]}{[\text{weak acid}]}\] We know that [weak acid] > [conjugate base], so the ratio \(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\) is less than 1. Since the logarithm of a value less than 1 is negative (log x < 0 if 0 < x < 1), we have: \[ pH = pKa + log(\text{negative value})\] Therefore, when [weak acid] > [conjugate base], the pH of the solution is less than the pKa value.
03

Determine the relationship between pH and pKa when [conjugate base] > [weak acid]

Given the condition [conjugate base] > [weak acid], we will use the Henderson-Hasselbalch equation: \[ pH = pKa + log \frac{[\text{conjugate base}]}{[\text{weak acid}]}\] We know that [conjugate base] > [weak acid], so the ratio \(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\) is greater than 1. Since the logarithm of a value greater than 1 is positive (log x > 0 if x > 1), we have: \[ pH = pKa + log(\text{positive value})\] Therefore, when [conjugate base] > [weak acid], the pH of the solution is greater than the pKa value.

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Most popular questions from this chapter

What volumes of \(0.50 \mathrm{M} \mathrm{HNO}_{2}\) and \(0.50 \mathrm{M} \mathrm{NaNO}_{2}\) must be mixed to prepare \(1.00 \mathrm{~L}\) of a solution buffered at \(\mathrm{pH}=3.55\) ?

Consider the titration of \(80.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\), by \(0.400\) \(M\) HCl. Calculate the pH of the resulting solution after the following volumes of \(\mathrm{HCl}\) have been added. a. \(0.0 \mathrm{~mL}\) d. \(40.0 \mathrm{~mL}\) b. \(20.0 \mathrm{~mL}\) e. \(80.0 \mathrm{~mL}\) c. \(30.0 \mathrm{~mL}\)

Consider a solution formed by mixing \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}, 30.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HOCl}, 25.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{NaOH}\). \(25.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\), and \(10.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{KOH}\). Calculate the \(\mathrm{pH}\) of this solution.

A certain buffer is made by dissolving \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in some water. Write equations to show how this buffer neutralizes added \(\mathrm{H}^{+}\) and \(\overline{\mathrm{OH}}^{-}\).

Derive an equation analogous to the Henderson-Hasselbalch equation but relating \(\mathrm{pOH}\) and \(\mathrm{p} K_{\mathrm{b}}\) of a buffered solution composed of a weak base and its conjugate acid, such as \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4}^{+}\).
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