CHEM 1001 Chapter Notes - Chapter 17: Titration Curve, Equivalence Point, Conjugate Acid

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29 Jun 2018

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CHAPTER 17: Additional Aspects of Acid-Base Equilibria

17.1 Common-Ion Effect In Acid-Base Equilibria: suppression of ionization of

weak electrolyte caused by adding more of one of its ionization products.

Can exist from strong acid that produces H3O+.

Due to eqbm expression, if concentration of acid and conjugate base are

same, then Ka = [H3O+] and pH = pKa (half neutralization point of titration

curve)

17.2 Buffer Solutions: pH changes very slightly when acid or base is added.

Has to be weak acid & its conj. base or weak base & its conj. acid  base

reacts with H3O+ & acid reacts with OH- when acid/base is added.

Henderson-Hasselbalch equation: pH = pKa + log([conjugate base]/

[acid])

Buffer is effective if ratio of base/acid is between 0.1 & 10, and molarity of

each buffer component exceeds by Ka by at least factor of 100 (so if

simplying assumption is valid)

Find amount of acid/base needed in buffer by finding [H3O+] & plugging in

Ka expression.

Can find how buffer pH was affected using equation & ICE table & equation

again.

Buffer capacity: amount of acid/base tha can be added before pH changes by

more than 2 unit range (+/- 1 unit)

17.3 Acid-Base Indicators: its colour depends on pH of solution to which it’s

added. It’s usually a weak base or acid, & doesn’t affect the eqbm if added in small

amounts.

17.4 Neutralization Reactions and Titration Curves

Equivalence point: point at which both acid & base are consumed

completely. Point when indicator changes colour is endpoint.

Titration curve: graph of pH versus volume of titrant (solution in buret)

added.

Strength of acid doesn’t affect volume of base required to titrate to

equivalence point.

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Document Summary

17. 1 common-ion effect in acid-base equilibria: suppression of ionization of weak electrolyte caused by adding more of one of its ionization products. Can exist from strong acid that produces h3o+. Due to eqbm expression, if concentration of acid and conjugate base are same, then ka = [h3o+] and ph = pka (half neutralization point of titration curve) 17. 2 buffer solutions: ph changes very slightly when acid or base is added. Has to be weak acid & its conj. base or weak base & its conj. acid base reacts with h3o+ & acid reacts with oh- when acid/base is added. Henderson-hasselbalch equation: ph = pka + log([conjugate base]/ Buffer is effective if ratio of base/acid is between 0. 1 & 10, and molarity of each buffer component exceeds by ka by at least factor of 100 (so if simplying assumption is valid) Find amount of acid/base needed in buffer by finding [h3o+] & plugging in.

Related Questions

Â± pH Changes in Buffers

When a solution contains a weak acid and its conjugate base or a weak base and its conjugate acid, it will be a buffer solution. Buffers resist change in pH following the addition of acid or base. A buffer solution prepared from a weak acid (HA) and its conjugate base (Aâ) is represented as

HA(aq)âH+(aq)+Aâ(aq)

The buffer will follow Le ChÃ¢telier's principle. If acid is added, the reaction shifts to consume the added H+, forming more HA. When base is added, the base will react with H+, reducing its concentration. The reaction then shifts to replace H+ through the dissociation of HA into H+ and Aâ. In both instances, [H+] tends to remain constant.

The pH of a buffer is calculated by using the Henderson-Hasselbalch equation:

pH=pKa+log[Aâ][HA]

Part A

What is the pH of a buffer prepared by adding 0.405 mol of the weak acid HA to 0.609 mol of NaAin 2.00 L of solution? The dissociation constant Ka of HA is 5.66Ã10â7.

Express the pH numerically to three decimal places.

Hints

pH =

6.424

SubmitMy AnswersGive Up

Correct

Since both the acid and base exist in the same volume, we can skip the concentration calculations and use the number of moles in the Henderson-Hasselbalch equation to calculate the pH. The answer will be the same.

Part B

What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid.

Express the pH numerically to three decimal places.

Hints

pH =

6.165

All attempts used; correct answer displayed

Even though a significant amount of a strong acid was added, the pH of the buffer changed very little. If this amount of acid was added to 2.00 L of pure water, the pH would change from 7.000 to 1.125.

Part C

What is the pH after 0.195 mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base.

Express the pH numerically to three decimal places.

pH =

6.408

Incorrect; Try Again; 4 attempts remaining

Need help with part B and C