CH_8_5_Reactions_of_Acids_and_Bases

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Chapter 8 Acids and Bases
1
8.5
Reactions of Acids and Bases
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Acids and Metals
2
Acids react with metals
• such as K, Na, Ca, Mg, Al, Zn, Fe, and Sn
• to produce hydrogen gas and the salt of the metal
Molecular equations:
2K(s) + 2HCl(aq)  2KCl(aq) + H2(g)
metal
acid
salt
hydrogen gas
Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
metal
acid
salt
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
hydrogen gas
Copyright © 2012 by Pearson Education, Inc.
Learning Check
3
Write a balanced equation for the reaction of magnesium
metal with HCl(aq). Label the metal, acid, and salt.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Learning Check
4
Write a balanced equation for the reaction of magnesium
metal with HCl(aq).
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
metal
acid
salt
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Acids and Carbonates
5
Acids react
• with carbonates and hydrogen carbonates
• to produce carbon dioxide gas, a salt, and water
2HCl(aq) + CaCO3(s)  CO2(g) + CaCl2(aq) + H2O(l)
acid
carbonate
carbon
salt
water
dioxide
HCl(aq) + NaHCO3(s)  CO2(g) + NaCl (aq) + H2O(l)
acid
bicarbonate
carbon
salt
water
dioxide
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Learning Check
6
Write a balanced equation for the following reactions:
A. MgCO3(s) + HBr(aq) 
B. HCl(aq) + NaHCO3(aq) 
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
7
Write a balanced equation for the following reactions:
A. MgCO3(s) + 2HBr(aq)  MgBr2(aq) + CO2(g) + H2O(l)
carbonate
acid
salt
carbon water
dioxide
B. HCl(aq) + NaHCO3(aq)  NaCl(aq) + CO2(g) + H2O(l)
acid
bicarbonate
salt
carbon water
dioxide
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Neutralization Reactions
8
In a neutralization reaction,
• an acid reacts with a base to produce salt and water
• the acid HCl reacts with NaOH to produce salt and water
• the salt formed is the anion from the acid and cation of
the base
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
acid
base
salt
water
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Neutralization Reactions
9
In neutralization reactions,
• if we write the strong acid and strong base as ions, we
see that H+ reacts with OH− to form water, leaving the ions
Na+ and Cl in solution:
H+(aq) + Cl(aq) + Na+(aq) + OH(aq) 
Na+(aq) + Cl(aq) + H2O(l)
• the overall reaction is H3O+ from the acid and OH from
the base form water:
H+(aq) + OH(aq)  H2O(l)
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Guide for Balancing Neutralization
Reactions
10
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Balancing Neutralization Reactions
11
Write the balanced equation for the neutralization of
magnesium hydroxide and nitric acid.
Step 1 Write the reactants and products.
Mg(OH)2 + HNO3
Step 2 Balance the H+ in the acid with the OH in the base.
Mg(OH)2 + 2HNO3
Step 3 Balance the H2O with H+ and the OH.
Mg(OH)2 + 2HNO3  salt + 2H2O
Step 4 Write the salt from the remaining ions.
Mg(OH)2 + 2HNO3  Mg(NO3)2 + 2H2O
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Learning Check
12
Select the correct group of coefficients for each of the
following neutralization equations.
1. HCl(aq) + Al(OH)3(aq)  AlCl3(aq) + H2O(l)
A. 1, 3, 3, 1
B. 3, 1, 1, 1
C. 3, 1, 1, 3
2. Ba(OH)2(aq) + H3PO4(aq)  Ba3(PO4)2(s) + H2O(l)
A. 3, 2, 2, 2
B. 3, 2, 1, 6
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
C. 2, 3, 1, 6
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Solution
13
1. HCl(aq) + Al(OH)3(aq)  AlCl3(aq) + H2O(l)
Step 1 Write the reactants and products.
HCl + Al(OH)3
Step 2 Balance the H+ in the acid with the OH in the base.
3HCl + Al(OH)3
Step 3 Balance the H2O with H+ and the OH.
3HCl + Al(OH)3  salt + 3H2O
Step 4 Write the salt from the remaining ions.
3HCl(aq) + Al(OH)3(aq)  AlCl3(aq) + 3H2O(l)
The answer is C. 3, 1, 1, 3.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
14
2. Ba(OH)2(aq) + H3PO4(aq)  Ba3(PO4)2(s) + H2O(l)
Step 1 Write the reactants and products.
Ba(OH)2 + H3PO4
Step 2 Balance the H+ in the acid with the OH in the base.
3Ba(OH)2 + 2H3PO4
Step 3 Balance the H2O with H+ and the OH.
3Ba(OH)2 + 2H3PO4  salt + 6H2O
Step 4 Write the salt from the remaining ions.
3Ba(OH)2(aq) + 2H3PO4(aq)
 Ba3(PO4)2(s) + 6H2O(l)
The answer is B. 3, 2, 1, 6.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Basic Compounds in Antacids
15
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Learning Check
16
Write the neutralization reactions for stomach acid, HCl,
and the ingredients in Mylanta.
Mylanta: Al(OH)3 and Mg(OH)2
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
17
Write the neutralization reactions for stomach acid, HCl, and
the ingredients in Mylanta.
Mylanta: For Al(OH)3:
Step 1 Write the reactants and products.
Al(OH)3 + HCl
Step 2 Balance the H+ in the acid with the OH in the base.
Al(OH)3 + 3HCl
Step 3 Balance the H2O with H+ and the OH.
Al(OH)3 + 3HCl  salt + 3H2O
Step 4 Write the salt from the remaining ions.
Al(OH)3(aq) + 3HCl(aq)  AlCl3(aq) + 3H2O(l)
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
18
Write the neutralization reactions for stomach acid, HCl, and
the ingredients in Mylanta.
Mylanta: For Mg(OH)2:
Step 1 Write the reactants and products.
Mg(OH)2 + HCl
Step 2 Balance the H+ in the acid with the OH in the base.
Mg(OH)2 + 2HCl
Step 3 Balance the H2O with H+ and the OH.
Mg(OH)2 + 2HCl  salt + 2H2O
Step 4 Write the salt from the remaining ions.
2HCl(aq) + Mg(OH)2(aq)  MgCl2(aq) + 2H2O(l)
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Acid–Base Titration
19
Titration
• is a laboratory procedure
used to determine the
molarity of an acid
• uses a base such as NaOH
to neutralize a measured
volume of an acid
Base 
NaOH
Acid 
Solution
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Indicator
20
An indicator
• is added to the acid in the
flask
• causes the solution to change
color when the acid is
neutralized
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Endpoint of Titration
21
At the endpoint,
• the indicator gives the
solution a permanent
pink color
• the volume of the base
used to reach the endpoint
is measured
• the molarity of the acid
is calculated using the
neutralization equation for
the reaction
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Guide to Calculations for
Acid–Base Titrations
22
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Acid–Base Titration Calculations
23
What is the molarity of an HCl solution if 18.5 mL of
0.225 M NaOH are required to neutralize 10.0 mL of HCl?
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Step 1 State given and needed quantities.
Given: 18.5 mL of 0.225 M NaOH
10.0 mL of HCl
Needed: Molarity of HCl
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Acid–Base Titration Calculations
24
What is the molarity of an HCl solution if 18.5 mL of
0.225 M NaOH are required to neutralize 10.0 mL of HCl?
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Step 2 Write a plan to calculate molarity or volume.
molarity
NaOH
equation
coefficients
molarity
HCl
18.5 mL  L  moles NaOH  moles HCl  L HCl
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Acid–Base Titration Calculations
25
What is the molarity of an HCl solution if 18.5 mL of
0.225 M NaOH are required to neutralize 10.0 mL of HCl?
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Step 3 State equalities and conversion factors including
concentration.
1 L = 1000 mL
0.225 mole NaOH
1 L NaOH
and
1 mole HCl x
1 mole NaOH
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Acid–Base Titration Calculations
26
What is the molarity of an HCl solution if 18.5 mL of
0.225 M NaOH are required to neutralize 10.0 mL of HCl?
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Step 4 Set up the problem to calculate the needed quantity.
18.5 mL NaOH  1 L NaOH  0.225 mole NaOH
1000 mL NaOH
1 L NaOH
 1 mole HCl = 0.00416 mole of HCl
1 mole NaOH
MHCl = 0.00416 mole HCl = 0.416 M HCl
0.0100 L HCl
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Learning Check
27
Calculate the mL of 2.00 M H2SO4 required to neutralize
50.0 mL of 1.00 M KOH.
H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l)
A. 12.5 mL
B. 50.0 mL
C. 200. mL
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
28
Calculate the mL of 2.00 M H2SO4 required to neutralize
50.0 mL of 1.00 M KOH.
H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l)
Step 1 State given and needed quantities.
Given: 2.00 M H2SO4
50.0 mL of 1.00 M KOH
Needed: mL of H2SO4
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
29
Calculate the mL of 2.00 M H2SO4 required to neutralize
50.0 mL of 1.00 M KOH.
H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l)
Step 2 Write a plan to calculate molarity or volume.
molarity
KOH
equation
coefficients
molarity
H2SO4
50.0 mL  L  moles KOH  moles H2SO4  L H2SO4
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
30
Calculate the mL of 2.00 M H2SO4 required to neutralize
50.0 mL of 1.00 M KOH.
H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l)
Step 3 State equalities and conversion factors including
concentration.
1 L = 1000 mL
1 L H2SO4
and
2.00 mole H2SO4
1 mole H2SO4
2 mole KOH
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
31
Calculate the mL of 2.00 M H2SO4 required to neutralize
50.0 mL of 1.00 M KOH.
H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l)
Step 4 Set up the problem to calculate the needed quantity.
50.0 mL  1 L = 0.0500 L
103 mL
0.0500 L KOH  1.00 mole KOH  1 mole H2SO4
1 L KOH
2 mole KOH
 1 L H2SO4
 1000 mL = 12.5 mL.
2.00 mole H2SO4 1 L H2SO4
The answer is A.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
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