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 waterMilk

VinegarLemon juiceStomach acid

1 M HCl

Aci

dic

N

eutr

al

Bas

ic

pH of Common Substances

Timberlake, Chemistry 7th Edition, page 335

1.0 MHCl0

gastricjuice1.6

vinegar2.8

carbonated beverage3.0

orange3.5

apple juice3.8

tomato4.2

lemonjuice2.2 coffee

5.0

bread5.5

soil5.5

potato5.8

urine6.0

milk6.4

water (pure)7.0

drinking water7.2

blood7.4

detergents8.0 - 9.0

bile8.0

seawater8.5

milk of magnesia10.5

ammonia11.0

bleach12.0

1.0 MNaOH(lye)14.0

8 9 10 11 12 14133 4 5 621 70

acidic neutral basic[H+] = [OH-]

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 Concentrations

pH = 3

pH = 7

pH = 11

OH-

H3O+OH-

OH-H3O+

H3O+

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

acidicsolution

neutralsolution

basicsolution

co

nc

en

trat

ion

(m

ole

s/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+]

Acid Dissociation

monoproticmonoprotic

diproticdiprotic

polyproticpolyprotic

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

0.03 M 0.03 M 0.03 M

pH = - log [H+]

pH = - log [0.03M]

pH = 1.52e.g. HCl, HNO3

H2A(aq) 2 H1+(aq) + A2-(aq)

0.3 M 0.6 M 0.3 M

pH = - log [H+]

pH = - log [0.6M]

pH = 0.22e.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] concentration

Note: 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

L

mol 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

x g Mg(OH)2 = 2.5 x 10-5 mol Mg(OH)2 1 mol Mg(OH)2

= 0.00145 g Mg(OH)2

58 g Mg(OH)2