Acid Base Reactions 1. 2. 3. 4. 5. 6. 7. 8. Objectives: Calculate hydrogen ion and hydroxide ion concentrations Calculate pH values Explain the action of buffers Explain why indicators change color Describe the.

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Transcript Acid Base Reactions 1. 2. 3. 4. 5. 6. 7. 8. Objectives: Calculate hydrogen ion and hydroxide ion concentrations Calculate pH values Explain the action of buffers Explain why indicators change color Describe the.

Acid Base Reactions
1.
2.
3.
4.
5.
6.
7.
8.
Objectives:
Calculate hydrogen ion and hydroxide ion concentrations
Calculate pH values
Explain the action of buffers
Explain why indicators change color
Describe the relationship between neutralization reactions and
titrations
Explain the results of hydrolyses of salt solutions
Select appropriate indicators for titrations
Use normalities in solving titration problems
Created by C. Ippolito Jan 2007
Self-ionization of Water
•Even “pure” water has some ions present
H2O + H2O H3O+ + OHthe longer arrow - water is far greater in
number than hydronium ions
H2O
H+ + OH-
[ H  ] x [OH  ]
 K eq
[ H 2O ]
[ H  ] x [OH  ]  K eq x [ H 2O]
[ H  ] x [OH  ]  KW
 ion
product for water or Kw
 at 25o is 1.0 x 10-14 mole2/L2
[ H  ] x [OH  ]  1.0 x 1014 mole2 /L2
Created by C. Ippolito Jan 2007
Self-ionization of Water
Calculate molar concentration of ions


14
[ H ] x [OH ]  1.0 x 10
14
2
2
mole /L
x  x  1.0 x 10 mole /L
14
2
2
x  1.0 x 10 mole /L
2
14
2
2
x  1.0 x 10 mole /L
2
2
7
7
x  1.0 x 10 mole /L  1.0 x 10 M
Created by C. Ippolito Jan 2007
Sample Problem
•
A mass of 4.0 g of NaOH is dissolved in
water to form 500 mL of solution with a
temperature of 25oC. What is the
hydrogen ion concentration?
NaOH(s)  Na+(aq) + OH-(aq)
1. Find number of moles in 4.0 g
2. Find molar concentration of OH3. Use ion product for water constant to find molar
concentration of H+
Created by C. Ippolito Jan 2007
Calculations
4g
 .10 m ole NaOH
40g
.10 m ole
x
[OH ] 

500 m l 1000m l

[OH  ]  500 ml  x  .10 mole 1000ml
.10 m ole 1000m l

[OH ]  x 
500 m l
[OH  ]  x  .2 M
[ H  ] x (.20 M )  1.0 x 1014 mole2 / L2
14
2
2
1
.
0
x
10
mole
/
L
[H  ] 
 5 x 1014 M
.20 M
Created by C. Ippolito Jan 2007
The pH of a Solution
• [H+] in scientific notation
• pH simplifies this
– pH values are the negative logarithm in
base 10
• pH = -log [H+]
– if [H+] = 1x 10-12
• pH = -(log 10-12) = -(-12) = 12
• pH = 12
• measures the relative strength of
acids and bases
Created by C. Ippolito Jan 2007
pH Scale
• acids have pH values below 7
pH
[H+]
[0H-]
0
100 or 1
10-14
1
10-1
10-13
2
10-2
10-12
3
10-3
10-11
4
10-4
10-10
5
10-5
10-9
6
10-6
10-8
Created by C. Ippolito Jan 2007
pH Scale
• water is neutral it has a pH of 7
pH
[H+]
[0H-]
7
10-7
10-7
Created by C. Ippolito Jan 2007
pH Scale
• bases have pH values above 7
pH
[H+]
[0H-]
8
10-8
10-6
9
10-9
10-5
10
10-10
10-4
11
10-11
10-3
12
10-12
10-2
13
10-13
10-1
14
10-14
100 or 1
Created by C. Ippolito Jan 2007
Calculations
• Given [H+] find pH.
– [H+] = 1 x 10-5
• pH = -log [H+] = -log [1 x 10-5] = -[-5]
• pH = 5.00
– [H+] = 3.5 x 10-8
•
•
•
•
pH = -log [H+] = -log [3.5 x 10-8]
pH = -(log3.5 + log10-8)
pH = -log3.5 – (-8)
pH = -.54 + 8.00 = 7.46
Created by C. Ippolito Jan 2007
Calculations
• Given pH find [H+].
– pH = 3.00
•
•
•
•
pH = -log [H+]
3.00 = -log [H+]
-3.00 = log [H+]
1.0 x 10-3 M = [H+]
– pH = 9.55
•
•
•
•
pH = -log [H+]
-pH = log [H+]
-9.55 = log [H+] use antilog function 10x
2.82 x 10-10 M = [H+]
Created by C. Ippolito Jan 2007
Buffers
• mixtures of chemicals that make a
solution resist changes in its pH
– types of buffers:
• weak acid and one of its soluble salts
– acetic acid and sodium acetate
» HC2H3O2 ↔H+ + C2H3O2-
• weak base and one of its soluble salts
– ammonia and ammonium chloride
– when acid added
• excess negative ions from salt “absorbs” H+
to form more of the weak acid
– when a base added
• H+ “absorbs” OH- to form water releasing
more negative ions of salt
Created by C. Ippolito Jan 2007
Acid Base Indicators
• Indicators
– change color when in different pH solutions
Indicator
Below
Range
Above
methyl orange
red
3.2-4.4
orange
yellow
bromthymol blue
yellow
6.0-7.6
blue
blue
bromcresol green
yellow
3.8-5.4
green
blue
litmus
red
5.5-8.2
red
blue
phenolphthalein
colorless
8.2-10
colorless
pink
thymol blue
yellow
8.0-9.6
blue
blue
Created by C. Ippolito Jan 2007
Titration
• laboratory procedure to determine
concentration of acid/base
– a known concentration acid/base is
added to the unknown concentration
acid/base to neutralize the unknown
• Standard Solution
– the known concentration acid/base
• End Point (Equivalence Point)
– indicator color change “show” neutralization
– recording the volume of the “known”
needed to “neutralize” a fixed volume of
the “unknown” allows calculation of the
concentration of the “unknown”
Created by C. Ippolito Jan 2007
Titration Problems
• Sample Problem:
– 20.0 mL solution of strontium hydroxide
Sr(OH)2 with indicator is neutralized when 25
mL of 0.0500 M HCl is added. What is the
concentration of Sr(OH)2?
– TO SOLVE:
• write and balance equation to see mole
relationship
• find molar concentration using mole relationships
– moles of acid = volume of acid (Va) x molar concentration
of acid (Ma)
– moles of base = volume of base (Vb) x molar
concentration of base (Mb)
• if necessary convert ml to L
Created by C. Ippolito Jan 2007
Calculations
• Equation
– HCl + Sr(OH)2  H2O + SrCl2
– 2HCl + Sr(OH)2  2H2O + SrCl2
• Mole Relationship
– # of moles of acid is twice the # of moles of
base
– moles of acid = 2(moles of base)
• Concentration Equation
– VaMa = 2VbMb solve for Mb
• Known:
– Va = 25 mL = .0250 L
– Ma = 0.0500 M
– Vb = 20 mL = .0200 L
Created by C. Ippolito Jan 2007
Va M a
Mb 
2Vb
Calculations (con’t)
.025 L x .500 m ole
L
Mb 
2 x .02 L
M b  .0312m ole  .0312M
L
Created by C. Ippolito Jan 2007
Hydrolysis of Salts
• Hydrolysis
– reaction of a substance with water
• Salts
– ionic compounds
– from neutralization of acids/bases
• Neutral Salt Solution
– no reaction with water
• Acidic Salt Solution
– reacts with water to produce H+ ions
• Basic Salt
– reacts with water to produce OH- ions
Created by C. Ippolito Jan 2007
Hydrolysis of Salts (con’t)
• Salt of Strong Base and Weak Acid
– NaC2H3O2 (sodium acetate)
• formed in neutralization
– strong base (NaOH) with weak acid (HC2H3O2)
• in water
• NaC2H3O2  Na+ + C2H3O2– the Na+ does not attract OH- from water
– but the C2H3O2- does attract H+ from water
• this leads to a slight excess of OH• hydrolysis of this salt causes a basic
solution to form
Created by C. Ippolito Jan 2007
Hydrolysis of Salts (con’t)
• Salt of Weak Base and Strong Acid
– NH4Cl (ammonium chloride)
• formed in neutralization
– weak base (NH3) with strong acid (HCl)
• in water
• NH4Cl  NH4+ + Cl– the Cl- does not attract from H+ water
– but the NH4+ does donate a proton to water
» NH4+ + H2O  H3O+ + NH3
• this leads to a slight excess of H+
• hydrolysis of this salt causes a acidic
solution to form
Created by C. Ippolito Jan 2007
Gram Equivalent Mass
• gram equivalent mass of an acid
m ass ( g ) of 1 m oleof acid
num berof protonsdonated
• gram equivalent mass of an base
m ass ( g ) of 1 m oleof base
num berof protonsaccepted
Created by C. Ippolito Jan 2007
Normality
• makes acid base titration calculations
easier
• Normality
# of gram equivalentm assesof solute
liter of solution
– Problem: A solution of 500mL is made
with 98 g of H2SO4. What is its
normality?
1 gram equivalent 
N
98 g
 49 g
2
2 gram equivalents 4
  4.0 N
.5 L
1
Created by C. Ippolito Jan 2007