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Chapter 17: Problem 92

Mathematically prove that the \(\mathrm{pH}\) at the halfway point of a titration of a weak acid with a strong base (where the volume of added base is half of that needed to reach the equivalence point) is equal to \(\mathrm{p} K_{a}\) for the acid.

Short Answer

Expert verified
At the halfway point, \(\mathrm{pH} = \mathrm{p}K_{a}\) for the weak acid.

Step by step solution

01

Understand the Halfway Point in Titration

The halfway point in the titration of a weak acid with a strong base is where exactly half of the weak acid has been converted to its conjugate base. At this point, the concentration of the weak acid (\[HA\]) is equal to the concentration of its conjugate base (\[A^-\]).
02

Consider the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation relates \(\mathrm{pH}\) to \(\mathrm{p}K_{a}\) by the formula: \[\mathrm{pH} = \mathrm{p}K_{a} + \log\left(\frac{\left[A^-\right]}{\left[HA\right]}\right)\] Since at the halfway point, \([A^-] = [HA]\), the expression simplifies the Log term.
03

Simplify the Equation

Substitute \([A^-] = [HA]\) into the equation: \[\log\left(\frac{[A^-]}{[HA]}\right) = \log(1) = 0\] This results in the equation becoming: \[\mathrm{pH} = \mathrm{p}K_{a} + 0\]
04

Conclude the Result

From the simplification, it follows that \(\mathrm{pH} = \mathrm{p}K_{a}\) at the halfway point. Therefore, the \(\mathrm{pH}\) at the halfway point in a titration is equal to the \(\mathrm{p}K_{a}\) of the weak acid.

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Key Concepts

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

Weak Acid
In chemistry, a weak acid is an acid that partially dissociates into ions in an aqueous solution. Unlike strong acids, which completely ionize, weak acids only partially release their hydrogen ions (H\(^+\)) when dissolved in water. This partial ionization is why weak acids are less conductive and generally have a higher pH compared to strong acids.
A classic example of a weak acid is acetic acid (CH\(_3\)COOH), commonly found in vinegar.
  • Weak acids have a corresponding acid dissociation constant, known as \(K_a\). The value of \(K_a\) indicates the strength of the acid; a smaller \(K_a\) value means a weaker acid.
  • The formula for \(K_a\) is: \[K_a = \frac{[H^+][A^-]}{[HA]}\]where \([HA]\) is the concentration of the undissociated form of the acid, and \([A^-]\) and \([H^+]\) are the concentrations of the acid's conjugate base and hydrogen ion, respectively.
Understanding these properties helps in predicting how the acid behaves in chemical reactions and solutions.
Strong Base
A strong base is a chemical compound that can completely dissociate into its ions within a solution. Unlike a weak base, a strong base releases all of its hydroxide ions (OH\(^-\)) into the solution, making it very efficient at neutralizing acids.
Common examples of strong bases include sodium hydroxide (NaOH) and potassium hydroxide (KOH).
  • In a strong base's dissociation, the reaction is represented as: \[\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-\]where the base breaks down completely in the aqueous solution.
  • Because they dissociate fully, solutions of strong bases tend to have a high pH, typically above 12.
When used in titration, a strong base will add significant OH\(^-\) ions to the solution, which helps in calculating and reaching the equivalence point.
Titration
Titration is a laboratory technique used to determine the concentration of a solute in a solution. By adding a titrant of known concentration to a solution with an unknown concentration, the titration reaction can help determine the unknown's concentration.
When a weak acid is titrated with a strong base, the point where exactly half of the acid has reacted to form its conjugate base is called the halfway point. At this halfway point:
  • The concentration of the acid equilibrates with its conjugate base, signifying that \[[HA] = [A^-]\]where \([HA]\) is the weak acid concentration and \([A^-]\) is its conjugate base concentration.
  • This property ensures that the Log term in the Henderson-Hasselbalch equation becomes zero, simplifying the pH calculation at this point.
Titration is a critical tool for assessing a chemical's property, such as its strength and behavior under chemical reactions.
Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is a vital formula in chemistry for calculating the pH of buffer solutions. This equation provides a relationship between the pH of a solution and the pKa of the acid with its conjugate base concentration.
The formula can be expressed as:\[\text{pH} = \text{p}K_a + \log\left(\frac{[A^-]}{[HA]}\right)\]
  • The equation highlights how the pH is influenced by the ratio of the concentrations of the conjugate base \([A^-]\) to the weak acid \([HA]\).
  • At the halfway point of a titration where \([A^-] = [HA]\), the logarithmic term becomes zero. As a result, the pH equals the pKa of the weak acid at this specific point.
This equation is especially useful in designing buffer solutions and understanding the behavior of acids and bases in various chemical environments.

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Most popular questions from this chapter

Which of these statements about the common-ion effect is most correct? (a) The solubility of a salt MA is decreased in a solution that already contains either \(\mathrm{M}^{+}\) or \(\mathrm{A}^{-}\). (b) Common ions alter the equilibrium constant for the reaction of an ionic solid with water. \((\mathbf{c})\) The common-ion effect does not apply to unusual ions like \(\mathrm{SO}_{3}^{2-}\). (d) The solubility of a salt MA is affected equally by the addition of either \(\mathrm{A}\) - or a noncommon ion.

Suppose you want to do a physiological experiment that calls for a pH 6.50 buffer. You find that the organism with which you are working is not sensitive to the weak acid \(\mathrm{H}_{2} \mathrm{~A}\left(K_{a 1}=2 \times 10^{-2} ; K_{a 2}=5.0 \times 10^{-7}\right)\) or its sodium salts. You have available a \(1.0 \mathrm{M}\) solution of this acid and \(\mathrm{a}\) \(1.0 \mathrm{M}\) solution of \(\mathrm{NaOH}\). How much of the \(\mathrm{NaOH}\) solution should be added to \(1.0 \mathrm{~L}\) of the acid to give a buffer at \(\mathrm{pH}\) \(6.50 ?\) (Ignore any volume change.)

A weak monoprotic acid is titrated with \(0.100 \mathrm{M} \mathrm{NaOH}\). It requires \(25.0 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution to reach the equivalence point. After \(12.5 \mathrm{~mL}\). of base is added, the pH of the solution is 4.16 . Estimate the \(\mathrm{p} K_{a}\) of the weak acid.

Compare the titration of a strong, monoprotic acid with a strong base to the titration of a weak, monoprotic acid with a strong base. Assume the strong and weak acid solutions initially have the same concentrations. Indicate whether the following statements are true or false. (a) More base is required to reach the equivalence point for the strong acid than the weak acid. (b) The \(\mathrm{pH}\) at the beginning of the titration is lower for the weak acid than the strong acid. \((\mathbf{c})\) The pH at the equivalence point is 7 no matter which acid is titrated.

(a) Calculate the percent ionization of \(0.250 \mathrm{M}\) lactic acid \(\left(K_{a}=1.4 \times 10^{-4}\right) .(\mathbf{b})\) Calculate the percent ionization of \(0.250 \mathrm{M}\) lactic acid in a solution containing \(0.050 \mathrm{M}\) sodium lactate.
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