Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Show 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 of a titration of a weak acid with a strong base, half of the weak acid has been neutralized by the strong base. Therefore, the concentrations of the weak acid HX and its conjugate base X- are equal at this point: \([HX] = [X^-]\). Plugging the concentrations of HX and X- into the Ka expression, we get \(K_a = [H^+]\). Taking the negative logarithm of both sides, we find that \(\mathrm{pKa} = \mathrm{pH}\). This demonstrates that the pH at the halfway point of a titration of a weak acid with a strong base is equal to the pKa for the acid.

Step by step solution

01

Define weak acids, strong bases, pH, and Ka

A weak acid is an acid that doesn't completely ionize in solution, meaning that not all of its molecules dissociate into H+ ions. A strong base, on the other hand, is a base that completely ionizes in solution, releasing OH- ions. The pH of a solution is a measure of its acidity or alkalinity, defined as pH = -log10[H+] where [H+] is the concentration of H+ ions in the solution. The pKa of an acid is the negative log of its Ka value, where Ka is the acid dissociation constant.
02

Explain the titration process and the equivalence point

Titration is an analytical technique used to determine the concentration of a solute in a solution. In the case of titrating a weak acid with a strong base, the weak acid is the solute, and the strong base is added gradually to the solution. The process continues until the number of moles of base added equals the initial number of moles of the acid, which is called the equivalence point.
03

Write the dissociation equation of the weak acid

In order to analyze the halfway point of the titration, we need the dissociation equation of the weak acid, HX, which can be written as: \[HX \rightleftharpoons H^+ + X^-\]
04

Determine the dissociation constants

The dissociation constant for the weak acid can be expressed as Ka: \[K_a = \frac{[H^+][X^-]}{[HX]}\] and the pH is related to the concentration of hydrogen ions [H+]: \[\mathrm{pH} = -\log_{10} [H^+]\]
05

Analyze the halfway point of the titration

At the halfway point of the titration, half of the weak acid has been neutralized by the strong base. Therefore, the concentrations of HX and its conjugate base X- are equal at this point: \[[HX] = [X^-]\]
06

Show that pH is equal to pKa at the halfway point

At the halfway point, we can plug the concentrations of HX and X- into the Ka expression: \[K_a = \frac{[H^+][X^-]}{[HX]}\] Since [HX] = [X-]: \[K_a = [H^+]\] Now, we can take the negative logarithm of both sides: \[-\log_{10}{K_a} = -\log_{10}{[H^+]}\] This gives us the relation between the pKa and the pH at the halfway point: \[\mathrm{pKa} = \mathrm{pH}\] This shows that the pH at the halfway point of a titration of a weak acid with a strong base is equal to the pKa for the acid, as required.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Weak Acid
A weak acid is a type of acid that does not completely dissociate in solution. This means that, in a solution of a weak acid, only a fraction of its molecules will give away their hydrogen ions (H⁺). This differs from strong acids, which dissociate almost completely in an aqueous solution.
For instance, acetic acid is a common example of a weak acid often used in chemistry. In water, acetic acid separates into acetate ions and hydrogen ions, but not fully, leaving some molecules intact. Here’s the dissociation reaction:\[ CH_3COOH ightleftharpoons CH_3COO^- + H^+ \]
  • Weak acids have higher pKa values compared to strong acids.
  • The extent of dissociation is determined by the acid's strength and the pH of the environment.
Understanding the behavior of weak acids is fundamental when studying acid-base titration because their dissociation is gradual, which influences the pH changes during the process.
Strong Base
A strong base is a substance that shows complete dissociation in water, breaking into its ions. This means that when a strong base is in solution, it releases all available hydroxide ions (OH⁻) into the solution. This property stands in contrast to weak bases, which only partially dissociate.
Sodium hydroxide (NaOH) is a classic example of a strong base. When dissolved in water, it fully separates into sodium (Na⁺) and hydroxide ions:\[ NaOH ightarrow Na^+ + OH^- \]
  • Strong bases raise the pH of a solution by a significant margin because they increase the OH⁻ ion concentration.
  • Knowing how strong bases behave is crucial when performing titrations, particularly titrations involving weak acids.
Utilizing a strong base in titration helps us determine the concentration of the solute by measuring the point at which the strong base neutralizes the weak acid.
Acid Dissociation Constant
The acid dissociation constant, Ka, is an essential concept in understanding how weak acids behave in solution. This constant provides a quantitative measure of the strength of an acid in solution, specifically describing how well an acid donates its protons to water, generating hydrogen ions. The equation for the Ka of a weak acid (HX) is:\[ K_a = \frac{[H^+][X^-]}{[HX]} \]
  • The larger the acid dissociation constant, the more the acid dissociates, indicating a stronger acid.
  • For weak acids, the Ka value is typically small, reflecting their lower tendency to lose H⁺ ions.
  • Understanding Ka is decisively important for calculating pH during titrations of weak acids by strong bases, especially at the halfway point.
The pKa, derived from Ka, is useful in theoretical and practical applications because it simplifies understanding by converting the value into a more manageable number. Knowing that at the halfway point of titration, pH equals pKa, makes for an important checkpoint in various chemical calculations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\begin{array}{llll} \text { (a) The molar solubility of } \mathrm{PbBr}_{2} & \text { at } 25^{\circ} \mathrm{C} & \text { is }\end{array}\) \(1.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L} .\) Calculate \(K_{s p} .(\mathbf{b})\) If \(0.0490 \mathrm{~g}\) of \(\mathrm{AgIO}_{3}\) dis- solves per liter of solution, calculate the solubility-product constant. (c) Using the appropriate \(K_{s p}\) value from Appendix \(\mathrm{D},\) calculate the \(\mathrm{pH}\) of a saturated solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\)

Explain why a mixture formed by mixing \(100 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(50 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\) will act as a buffer.

Predict whether the equivalence point of each of the following titrations is below, above, or at \(\mathrm{pH}\) 7: (a) \(\mathrm{NaHCO}_{3}\) titrated with \(\mathrm{NaOH},\) (b) \(\mathrm{NH}_{3}\) titrated with \(\mathrm{HCl}\) (c) KOH titrated with HBr.

(a) Explain the difference between solubility and solubilityproduct constant. (b) Write the expression for the solubility-product constant for each of the following ionic compounds: \(\mathrm{MnCO}_{3}, \mathrm{Hg}(\mathrm{OH})_{2},\) and \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

A buffer is prepared by adding \(20.0 \mathrm{~g}\) of sodium acetate \(\left(\mathrm{CH}_{3} \mathrm{COONa}\right)\) to \(500 \mathrm{~mL}\) of a \(0.150 \mathrm{M}\) acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) solution. (a) Determine the \(\mathrm{pH}\) of the buffer. (b) Write the complete ionic equation for the reaction that occurs when a few drops of hydrochloric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of sodium hydroxide solution are added to the buffer.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free