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

Short Answer

Expert verified
To sketch a pH curve for the titration of a weak acid (HA) with a strong base (NaOH), identify the major species at various points during the titration process and determine the pH using the Ka and Kw expressions. The major species at significant points are: HA (initial point), HA, A-, OH- (before equivalence point), HA, A- (halfway point), A-, H2O (equivalence point), and A-, OH- (after equivalence point). Use stoichiometry and Ka/Kb expressions to calculate the pH at each point: for the initial point and before equivalence point, use Ka; for the halfway point, -log(Ka); for the equivalence point, use Kb; and for after equivalence point, use the concentration of excess OH- ions. Lastly, plot the pH values on the y-axis and the volume of NaOH added on the x-axis to create the pH curve, showing the gradual increase in the buffering region and the steep rise around the equivalence point.

Step by step solution

01

Analyze the given problem

We are given a weak acid (HA) titrated with a strong base (NaOH). First, we should write down the chemical reactions and formulas for calculating the pH of the solution. Dissociation of HA: HA ⇌ H+ + A- Neutralization reaction: HA + OH- → H2O + A- We will use the dissociation constant (Ka) and the ion-product constant of water (Kw) to calculate the pH. Ka = [H+][A-]/[HA] Kw = [H+][OH-]
02

Major species at various points

At different points during a titration process, we will have various major species present in the solution. Here, we will identify these major species at each significant point. 1. Initial point (no base added): HA (only the weak acid) 2. Before equivalence point (less base added): HA, A-, and OH- 3. At the halfway point (halfway to equivalence point): HA and A- 4. Equivalence point (equal amounts of acid and base): A- and H2O 5. After equivalence point (more base added): A- and OH-
03

Calculating the pH

With the major species identified, we'll calculate the pH at various points during the titration process. 1. Initial point: As no base has been added, we will use the Ka expression to calculate the [H+], then the pH. 2. Before equivalence point: To determine the pH, we will use the stoichiometry of the titration reaction and the Ka expression. 3. Halfway point: Since we know that Ka = [H+], we can directly use the -log of Ka to find the pH. 4. Equivalence point: At this point, all the weak acid has reacted with the base and only the A- ion is left. We'll use the Kb expression and the concentration of A- to calculate the OH- concentration, then pH. Kb = Kw/Ka 5. After equivalence point: Since there is an excess of NaOH at this point, we will use the concentration of the excess OH- ions to find the pOH, then the pH.
04

Sketching the pH curve

With the pH values calculated for various points, we can now sketch the pH curve. On the x-axis, we will plot the volume of NaOH added, and on the y-axis, we will plot the pH of the solution. The curve starts with the initial pH of the weak acid, increases gradually in the buffering region, then rises steeply around the equivalence point before leveling off again once we are past the equivalence point.

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!

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

Lactic acid is a common by-product of cellular respiration and is often said to cause the "burn" associated with strenuous activity. A \(25.0-\mathrm{mL}\) sample of \(0.100 M\) lactic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}, \mathrm{p} K_{\mathrm{a}}=\right.\) \(3.86\) ) is titrated with \(0.100 M\) NaOH solution. Calculate the \(\mathrm{pH}\) after the addition of \(0.0 \mathrm{~mL}, 4.0 \mathrm{~mL}, 8.0 \mathrm{~mL}, 12.5 \mathrm{~mL}, 20.0 \mathrm{~mL}\), \(24.0 \mathrm{~mL}, 24.5 \mathrm{~mL}, 24.9 \mathrm{~mL}, 25.0 \mathrm{~mL}, 25.1 \mathrm{~mL}, 26.0 \mathrm{~mL}, 28.0 \mathrm{~mL}\) and \(30.0 \mathrm{~mL}\) of the \(\mathrm{NaOH}\). Plot the results of your calculations as pH versus milliliters of \(\mathrm{NaOH}\) added.

The active ingredient in aspirin is acetylsalicylic acid. A \(2.51-g\) sample of acetylsalicylic acid required \(27.36 \mathrm{~mL}\) of \(0.5106 M\) \(\mathrm{NaOH}\) for complete reaction. Addition of \(13.68 \mathrm{~mL}\) of \(0.5106 \mathrm{M}\) \(\mathrm{HCl}\) to the flask containing the aspirin and the sodium hydroxide produced a mixture with \(\mathrm{pH}=3.48\). Determine the molar mass of acetylsalicylic acid and its \(K_{\mathrm{a}}\) value. State any assumptions you must make to reach your answer.

A student dissolves \(0.0100 \mathrm{~mol}\) of an unknown weak base in \(100.0 \mathrm{~mL}\) water and titrates the solution with \(0.100 \mathrm{M} \mathrm{HNO}_{3}\). After \(40.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{3}\) was added, the \(\mathrm{pH}\) of the resulting solution was \(8.00 .\) Calculate the \(K_{\mathrm{b}}\) value for the weak base.

Calculate the \(\mathrm{pH}\) of each of the following buffered solutions. a. \(0.50 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} / 0.25 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\) b. \(0.25 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{NH}_{2} / 0.50 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\) c. \(0.50 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} / 0.50 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\)

Consider the following four titrations. i. \(100.0 \mathrm{~mL}\) of \(0.10 M \mathrm{HCl}\) titrated by \(0.10 \mathrm{M} \mathrm{NaOH}\) ii. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{NaOH}\) titrated by \(0.10 \mathrm{M} \mathrm{HCl}\) iii. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{2}\) titrated by \(0.10 \mathrm{M} \mathrm{HCl}\) iv. \(100.0 \mathrm{~mL}\) of \(0.10 M\) HF titrated by \(0.10 \mathrm{M} \mathrm{NaOH}\) Rank the titrations in order of: a. increasing volume of titrant added to reach the equivalence point. b. increasing pH initially before any titrant has been added. c. increasing \(\mathrm{pH}\) at the halfway point in equivalence. d. increasing \(\mathrm{pH}\) at the equivalence point. How would the rankings change if \(\mathrm{C}_{5} \mathrm{H}_{3} \mathrm{~N}\) replaced \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) and if \(\mathrm{HOC}_{6} \mathrm{H}_{5}\) replaced \(\mathrm{HF}\) ?

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