pH Scale : pH Scale Soren Sorensen
(1868 - 1939)
pH Scale : pH Scale Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 515
pH of Common Substances : pH of Common Substances Timberlake, Chemistry 7th Edition, page 335 1.0 M
HCl
0 gastric
juice
1.6 vinegar
2.8 carbonated
beverage
3.0 orange
3.5 apple juice
3.8 tomato
4.2 lemon
juice
2.2 coffee
5.0 bread
5.5 soil
5.5 potato
5.8 urine
6.0 milk
6.4 water (pure)
7.0 drinking water
7.2 blood
7.4 detergents
8.0 - 9.0 bile
8.0 seawater
8.5 milk of
magnesia
10.5 ammonia
11.0 bleach
12.0 1.0 M
NaOH
(lye)
14.0 8 9 10 11 12 14 13 3 4 5 6 2 1 7 0 acidic neutral basic [H+] = [OH-]
pH of Common Substance : 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 M
Household bleach
Household ammonia
Lime water
Milk of magnesia
Borax
Baking soda
Egg white, seawater
Human blood, tears
Milk
Saliva
Rain
Black coffee
Banana
Tomatoes
Wine
Cola, vinegar
Lemon juice
Gastric juice More basic More acidic pH [H1+] [OH1-] pOH 7 1 x 10-7 1 x 10-7 7
Acid – Base Concentrations : Acid – Base Concentrations pH = 3 pH = 7 pH = 11 OH- H3O+ OH- OH- H3O+ H3O+ [H3O+] = [OH-] [H3O+] > [OH-] [H3O+] < [OH-] acidic
solution neutral
solution basic
solution concentration (moles/L) 10-14 10-7 10-1 Timberlake, Chemistry 7th Edition, page 332
pH : pH pH = -log [H1+] Kelter, Carr, Scott, Chemistry A World of Choices 1999, page 285
pH Calculations : 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 = - log [H+] pH = 4.6 pH = - log [H+] 4.6 = - log [H+] - 4.6 = log [H+] - 4.6 = log [H+] Given: 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+]
pH and pOH Calculations : pH and pOH Calculations Keys pH and pOH Calculations pH and pOH Calculations http://www.unit5.org/chemistry/AcidBase.html
Acid Dissociation : Acid Dissociation monoprotic diprotic polyprotic HA(aq) H1+(aq) + A1-(aq) 0.03 M 0.03 M 0.03 M pH = - log [H+] pH = - log [0.03M] pH = 1.52 e.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 = 0.22 e.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 = ?
Slide 11 : 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+] [H+] = 10-pH [H+] = 10-2.1 [H+] = 0.00794 M [H+] = 7.94 x10-3 M 7.94 x10-3 M Step 3) Calculate [H3PO4] concentration Note: coefficients (1:3) for (H3PO4 : H+) = 0.00265 M H3PO4
Slide 12 : 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)2 Mg(OH)2(aq) Mg2+(aq) 2 OH1-(aq) + Step 2) Calculate the pOH pH + pOH = 14 10.0 + pOH = 14 pOH = 4.0 Step 3) Calculate the [OH1-] pOH = - log [OH1-] [OH1-] = 10-OH [OH1-] = 1 x10-4 M 1 x10-4 M 0.5 x10-4 M 5 x10-5 M Step 4) Solve for moles of Mg(OH)2 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
Practice Problems - Key : Practice Problems - Key Keys Practice Problems - Answer Key Practice Problems - Answer Key http://www.unit5.org/chemistry/AcidBase.html