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[H + ] 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 10 0 0. 1 M NaOH. Ammonia (household cleaner). 7. - PowerPoint PPT Presentation
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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
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