Transcript Reactions of Acids & Bases

Chapter 8 Acids and Bases
8.6
Reactions of Acids and Bases
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
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Acids and Metals
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)
Zn(s) + 2HCl(aq)
ZnCl2(aq) + H2(g)
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Acids and Carbonates
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)
HCl(aq)
CO2(g) + NaCl (aq) + H2O(l)
+ NaHCO3(s)
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Learning Check
Write the products of the following reactions of acids.
A. Zn(s) + 2 HCl(aq)
B. MgCO3(s) + 2HCl(aq)
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Solution
Write the products of the following reactions of acids.
A. Zn(s) + 2 HCl(aq)
B. MgCO3(s) + 2 HCl(aq)
ZnCl2(aq) + H2(g)
MgCl2(aq) + CO2(g) + H2O(l)
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Neutralization Reactions
In a neutralization reaction
• an acid such as HCl reacts with a base such as NaOH.
HCl + H2O
H3O+ + Cl−
NaOH
Na+ + OH−
• the H3O+ from the acid and the OH− from the base form
water.
H3O+ + OH−
2 H2O
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Neutralization Equations
In the equation for neutralization, an acid and a base
produce a salt and water.
acid
base
salt
water
HCl + NaOH
NaCl
+ H 2O
2HCl + Ca(OH)2
CaCl2
+ 2H2O
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Balancing Neutralization Reactions
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Balancing Neutralization Reactions
Write the balanced equation for the neutralization of
magnesium hydroxide and nitric acid.
STEP 1 Write the acid and base.
Mg(OH)2 + HNO3
STEP 2 Balance H+ in acid with OH- in base.
Mg(OH)2+ 2HNO3
STEP 3 Balance with H2O.
Mg(OH)2 + 2HNO3
salt + 2H2O
STEP 4 Write the salt from remaining ions.
Mg(OH)2 + 2HNO3
Mg(NO3)2 + 2H2O
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Learning Check
Select the correct group of coefficients for each of the
following neutralization equations
A. HCl (aq) + Al(OH)3(aq)
AlCl3(aq) + H2O(l)
1) 1, 3, 3, 1
2) 3, 1, 1, 1
3) 3, 1, 1 3
B. Ba(OH)2(aq) + H3PO4(aq)
Ba3(PO4)2(s) + H2O(l)
1) 3, 2, 2, 2
2) 3, 2, 1, 6
3) 2, 3, 1, 6
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Solution
A. 3) 3, 1, 1 3
3HCl(aq + Al(OH)3(aq)
AlCl3(aq) + 3H2O(l)
B. 2) 3, 2, 1, 6
3Ba(OH)2 (aq) + 2H3PO4(aq)
Ba3(PO4)2(s)+ 6H2O(l)
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Basic Compounds in Some Antacids
Antacids are used to neutralize stomach acid (HCl).
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Learning Check
Write the neutralization reactions for stomach acid
HCl and Mylanta.
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Solution
Write the neutralization reactions for stomach acid
HCl and Mylanta.
Mylanta: Al(OH)3 and Mg(OH)2
3HCl(aq) + Al(OH)3(aq)
AlCl3(aq) + 3H2O(l)
2HCl(aq) + Mg(OH)2(aq)
MgCl2(aq) + 2H2O(l)
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Acid-Base Titration
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
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
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Indicator
An indicator
• is added to the acid in
the flask.
• causes the solution to
change color when the
acid is neutralized.
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
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End Point of Titration
At the end point,
• the indicator gives the solution a
permanent pink color.
• the volume of the base used to
reach the end point is measured.
• the molarity of the acid is
calculated using the neutralization
equation for the reaction.
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
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Calculating Molarity
What is the molarity of an HCl solution if 18.5 mL of a
0.225 M NaOH are required to neutralize 10.0 mL HCl?
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l)
STEP 1 Given: 18.5 mL of 0.225 M NaOH; 10.0 mL HCl
Need: Molarity of HCl
STEP 2 18.5 mL
L
moles NaOH
moles HCl M HCl
L HCl
STEP 3 1 L = 1000 mL
0.225 mole NaOH/1 L NaOH
1 mole HCl/1 mole NaOH
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Calculating Molarity (continued)
STEP 4 Calculate the molarity of HCl.
18.5 mL NaOH x 1 L NaOH
x 0.225 mole NaOH
1000 mL NaOH
1 L NaOH
L
x 1 mole HCl
1 mole NaOH
moles NaOH
=
0.00416 mole HCl
MHCl = 0.00416 mole HCl = 0.416 M HCl
0.0100 L HCl
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Learning Check
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)
1) 12.5 mL
2) 50.0 mL
3) 200. mL
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Solution
1) 12.5 mL
0.0500 L KOH x 1.00 mole KOH x 1 mole H2SO4
1 L KOH
2 mole KOH
1 L H2SO4
2.00 mole H2SO4
x
1000 mL =
1 L H2SO4
x
12.5 mL
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Learning Check
A 25.0 mL sample of phosphoric acid is neutralized by
42.6 mL of 1.45 M NaOH. What is the molarity of the
phosphoric acid solution?
3NaOH(aq) + H3PO4 (aq)
Na3PO4(aq) + 3H2O(l)
1) 0.620 M
2) 0.824 M
3) 0.185 M
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Solution
2) 0.824 M
0.0426 L x 1.45 mole NaOH x 1 mole H3PO4
1 L
3 mole NaOH
= 0.0206 mole H3PO4
0.0206 mole H3PO4 = 0.824 mole/L = 0.824 M
0.0250 L
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