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Chapter 10

Monoprotic Acid and Bases

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Strong Acids and Bases

Compound that when dissolved in water will fully dissociate.

This is a factor of our very universal solvent and its’ special properties. This is called the leveling effect of water.

Our strong acids are HCl HBr HI HNO3 HClO4 H2SO4 (first ionization)

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Strong Acids

The acid dissociation equilibrium HA + H2O = H3O+ + A-

The equilibrium constant values for our strong acids:

Acid Ka

HCl 7900HBr 630 000HI 25 000 000 000 HNO3 25

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Strong Acids

The strongest of the common strong acids is perchloric acid. (HClO4) This strengths can be shown be going to

other solvents other than water. Acidic acid is a common choice.

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Strong Acid and Bases

Our strong bases are NaOH KOH LiOH Sr(OH)2

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Determining pH

Since they completely dissociate we get the same concentration of H+ as the dissolved acid or base. So if a solution is made up to be 1.00x10-2 M

HCl then the pH will be pH = - log [H+] = -log (1.00x10-2) = 2.?????? (How many sig figs?)

More correctly we should correct for activity so we should have pH = -logAH+ = -log [H+] = -log (0.914)(1.00x10-2) = 2.03905 (How many sig figs?)

But we will not bother to correct unless called on to do so.

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Strong Acids

pH of strong base solutions. pH of a 1.0x10-3 M NaOH solution.

[Na+] = .0010 and [OH-] = 0.0010 Then pOH is 3.00

Since [H+][OH-] = Kw

pH + pOH = 14.00 Then pH of this solution is 11.00 Kw varies with temperature so you need to account

for that also.

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Strong Acids

What if we make up a solution that is 1.0x10-5

M HNO3. The pH would be 5.00. If we were to dilute this solution by 1000 fold.

That is take 1.00 mL and dilute to 1.00 Liters. What is the pH then.

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Strong Acids

It can not be pH 8. That would mean that we have made up a solution that is basic from and acid and water???

How would you calculate this?

This is a case for Chapter 9. Systematic treatment of equilibrium.

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Strong Acids

What do we know? CHNO3 = 1.00x10-8

From this we can see that CNO3- = 1.00x10-8

What is the charge balance of this sytem. [H+] = [NO3

-] + [OH-]

We also have our Kw expression. So [H+] = CNO3 + Kw/[H+]

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Weak Acids/Bases

These are acids and bases that do not fully dissociate when placed in aqueous solution.

Examples Acetic Acid Benzoic Acid Hydrofluoric Acid Ammonia

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Weak Acids/Bases

Weak acid equilibrium

HA = H+ + A-

][

]][[

HA

AHKa

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Weak Acids and Bases

Weak Base equlibrium (Also called hydrolysis constant)

B + H2O = BH+ + OH-

][

]][[

B

OHBHKb

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Conjugate Pairs

OH

O

H

O-

O

H

Formic Acid Formate Ion

An acid / conjugate base pair

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Conjugates

N

H

H H

N+

H

HH

H

Ammonia Ammonium IonBase Conjugate Acid

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Table G

The Table in the back of the book. Starting on page AP12 lists all acids and bases as acids. It will give the proper base name but show the structure for the conjugate acid form. When there is more than one acidic group then the book will indicate which proton goes with the given pKa.

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Table G

N+

OH H

8- Hydroxyquinoline

4.91 (NH)

9.81 (OH)

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Buffers

When you add a weak acid and its’ conjugate then you will get a buffer. Buffers resist the rapid change in pH when acid or base is added to the solution.

pH = pKa + log (base/acid) pH control is important since many

processes are pH dependent.

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Buffers

Made two different ways Mix an acid and its conjugate base

Acetic acid and sodium acetate Ammonium Chloride and aqueous

ammonium Prepare a solution of an acid or base

and generate the conjugate by addition of strong acid or base.

Acetic acid and add NaOH

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Buffers

When adding both forms the concentration of the buffer will be the sum of the concentration of both forms

When generating from strong conjugate then add the moles of acid needed at the final volume.

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Preparation

Weight out the number of moles of buffer needed and dilute to ~80% final volume.

Place pH electrode into solution Add strong acid/base until pH is

reached. Dilute to mark

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Buffer Capacity

The amount of acid or base that can be added before the buffer is consumed.

= dCb / dpH = - dCa / dpH

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Buffer pH

Ionic Strength

Temperature

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What about extreme conditions

FHA + FA- = [HA] + [A-] [Na+] + [H+] = [OH-] + [A-]

[HA] = FHA – [H+] + [OH-] [A-] = FA- + [H+] – [OH-]

pH = pKa + log {corr A- / corr HA}

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