pH Scale

Soren Sorensen(1868 - 1939)

pH Scale

Acid Base

0

7

14

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 515

[H[H++] pH] pH

10-14 14

10-13 13

10-12 12

10-11 11

10-10 10

10-9 9

10-8 8

10-7 7

10-6 6

10-5 5

10-4 4

10-3 3

10-2 2

10-1 1

100 0

1 M NaOH

Ammonia(householdcleaner)

BloodPure wateMilk

VinegarLemon juiceStomach acid

1 M HCl

Aci

dic

N

eutra

l

Bas

ic

pH of Common Substances

Timberlake, Chemistry 7th Edition, page 335

pH of Common Substance

14 1 x 10-14 1 x 10-0 0 13 1 x 10-13 1 x 10-1 1 12 1 x 10-12 1 x 10-2 2 11 1 x 10-11 1 x 10-3 3 10 1 x 10-10 1 x 10-4 4 9 1 x 10-9 1 x 10-5 5 8 1 x 10-8 1 x 10-6 6

6 1 x 10-6 1 x 10-8 8 5 1 x 10-5 1 x 10-9 9 4 1 x 10-4 1 x 10-10 10 3 1 x 10-3 1 x 10-11 11 2 1 x 10-2 1 x 10-12 12 1 1 x 10-1 1 x 10-13 13 0 1 x 100 1 x 10-14 14

NaOH, 0.1 MHousehold bleachHousehold ammonia

Lime waterMilk of magnesia

Borax

Baking sodaEgg white, seawaterHuman blood, tearsMilkSalivaRain

Black coffeeBananaTomatoesWineCola, vinegarLemon juice

Gastric juice

Mor

e ba

sic

Mor

e ac

idic

pH [H1+] [OH1-] pOH

7 1 x 10-7 1 x 10-7 7

Acid – Base ConcentrationspH = 3

pH = 7

pH = 11

OH-

H3O+OH-

OH-H3O+

H3O+

[H3O+] = [OH-] [H3O+] > [OH-] [H3O+] < [OH-] acidic

solutionneutralsolution

basicsolution

conc

entr

atio

n (m

oles

/L)

10-14

10-7

10-1

Timberlake, Chemistry 7th Edition, page 332

pH

pH = -log [H1+]

Kelter, Carr, Scott, Chemistry A World of Choices 1999, page 285

pH Calculations

pH

pOH

[H3O+]

[OH-]

pH + pOH = 14

pH = -log[H3O+]

[H3O+] = 10-pH

pOH = -log[OH-]

[OH-] = 10-pOH

[H3O+] [OH-] = 1 x10-14

pH = - log [H+]pH = 4.6

pH = - log [H+]

4.6 = - log [H+]

- 4.6 = log [H+]

- 4.6 = log [H+]

Given:

2nd log

10x

antilog

multiply both sides by -1

substitute pH value in equation

take antilog of both sides

determine the [hydronium ion]

choose proper equation

[H+] = 2.51x10-5 M

You can check your answer by working backwards.

pH = - log [H+]

pH = - log [2.51x10-5 M]

pH = 4.6

Recall, [H+] = [H3O+]

http://www.unit5.org/christjs/tempT27dFields-Jeff/AcidBase1.htm

pH and pOH Calculations

        pH and pOH Calculations

        pH and pOH Calculations

Acid Dissociation

monoproticmonoprotic

diproticdiprotic

polyproticpolyprotic

HA(aq) H1+(aq) + A1-(aq) 0.3 M 0.3 M 0.3 M

pH = - log [H+]pH = - log [3M]

pH = - 0.48e.g. HCl, HNO3

H2A(aq) 2 H1+(aq) + A2-(aq) 0.3 M 0.6 M 0.3 M

pH = - log [H+]pH = - log [6M]

pH = - 0.78e.g. H2SO4

Given: pH = 2.1

find [H3PO4]

assume 100% dissociation

e.g. H3PO4

H3PO4(aq) 3 H1+(aq) + PO43-(aq)

? M x M

pH = ?

Given: pH = 2.1

find [H3PO4]

assume 100% dissociation

H3PO4(aq) 3 H1+(aq) + PO43-(aq)

X M 0.00794 M

Step 1) Write the dissociation of phosphoric acid

Step 2) Calculate the [H+] concentration pH = - log [H+]

2.1 = - log [H+]

- 2.1 = log [H+]

2nd log - 2.1 = log [H+]2nd log

[H+] = 10-pH

[H+] = 10-2.1

[H+] = 0.00794 M

[H+] = 7.94 x10-3 M7.94 x10-3 M

Step 3) Calculate [H3PO4] concentrationNote: coefficients (1:3) for (H3PO4 : H+)

7.94 x10-3 M3

= 0.00265 M H3PO4

How many grams of magnesium hydroxide are needed to add to 500 mL of H2O to yield a pH of 10.0?

Step 1) Write out the dissociation of magnesium hydroxide Mg2+ OH1-

Mg(OH)2Mg(OH)2(aq) Mg2+(aq) 2 OH1-(aq)+

Step 2) Calculate the pOH pH + pOH = 1410.0 + pOH = 14

pOH = 4.0

Step 3) Calculate the [OH1-] pOH = - log [OH1-]

[OH1-] = 10-OH

[OH1-] = 1 x10-4 M

1 x10-4 M0.5 x10-4 M5 x10-5 M

Step 4) Solve for moles of Mg(OH)2

Lmol M L 0.5

molx M x105 5- x = 2.5 x 10-5 mol Mg(OH)2

Step 5) Solve for grams of Mg(OH)2

22

22

5-2 Mg(OH) g 0.00145

Mg(OH) mol (1Mg(OH) g (58Mg(OH) mol 10 x 2.5 Mg(OH) gx

))

http://www.unit5.org/christjs/tempT27dFields-Jeff/AcidBase1.htm

Practice Problems - Key

        Practice Problems - Answer Key

        Practice Problems - Answer Key