Transcript Electrolytes - College of San Mateo

Chapter 10
Acids and Bases
10.5
Reactions of Acids and Bases
General, Organic, and Biological Chemistry
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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
Equations:
2K(s) + 2HCl(aq)
2KCl(aq) + H2(g)
Zn(s) + 2HCl(aq)
ZnCl2(aq) + H2(g)
Net ionic equations:
2K(s) + 2H+(aq)
Zn(s) + 2H+(aq)
General, Organic, and Biological Chemistry
2K+(aq) + H2(g)
Zn2+ (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)
HCl(aq) + NaHCO3(s)
General, Organic, and Biological Chemistry
CO2(g) + CaCl2(aq) + H2O(l)
CO2(g) + NaCl (aq) + H2O(l)
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Learning Check
Write the products of the following reactions of acids
as the complete equation and net ionic equation:
A. Zn(s) + 2 HCl (aq)
B. MgCO3 (s) + 2HCl(aq)
General, Organic, and Biological Chemistry
?
?
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Solution
Write the products of the following reactions of acids
as the complete equation and net ionic equation:
A. Zn(s) + 2HCl(aq)
ZnCl2(aq) + H2(g)
Zn(s) + 2H+(aq)
Zn2+(aq) + H2(g)
B. MgCO3(s) + 2HCl(aq)
MgCO3(s) + 2H+(aq)
General, Organic, and Biological Chemistry
MgCl2(aq) + CO2(g) + H2O(l)
Mg2+(aq) + H2(g)
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Neutralization Reactions
Neutralization is the reaction of
 an acid such as HCl and a base such as NaOH
HCl(aq) + H2O(l)
H3O+ (aq) + Cl−(aq)
NaOH(aq)
Na+ (aq) + OH−(aq)
 the H3O+ from the acid and the OH− from the base
to form water
H3O+(aq) + OH−(aq)
2H2O(l)
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Neutralization Equations
 In the equations for neutralization, an acid and a
base produce a salt and water.
acid
base
salt
water
HCl(aq) + NaOH(aq)
H+(aq) + OH−(aq)
NaCl(aq) + H2O(l)
H2O
2HCl(aq) + Ca(OH)2(aq)
H+(aq) + OH−(aq)
CaCl2(aq) + 2H2O(l)
H2O(l)
General, Organic, and Biological Chemistry
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Guide to Balancing an Equation
for Neutralization
General, Organic, and Biological Chemistry
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Balancing Neutralization
Reactions
Write the balanced equation for the neutralization of
magnesium hydroxide and nitric acid.
STEP 1 Write the base and acid formulas:
Mg(OH)2(aq) + HNO3(aq)
STEP 2 Balance OH– and H+:
Mg(OH)2(aq) + 2HNO3(aq)
STEP 3 Balance with H2O:
Mg(OH)2(aq) + 2HNO3(aq)
salt + 2H2O(l)
STEP 4 Write the salt from remaining ions:
Mg(OH)2(aq) + 2HNO3(aq)
Mg(NO3)2(aq) + 2H2O(l)
General, Organic, and Biological Chemistry
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Learning Check
Select the correct group of coefficients for 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)
General, Organic, and Biological Chemistry
Ba3(PO4)2(s) + 6H2O(l)
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Antacids
Antacids
 are used to neutralize stomach acid (HCl)
General, Organic, and Biological Chemistry
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Learning Check
Write the neutralization reactions for stomach acid
HCl and Mylanta.
General, Organic, and Biological Chemistry
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Solution
Write the neutralization reactions for stomach acid
HCl and Mylanta.
STEP 1 Mylanta: Al(OH)3 and Mg(OH)2
Write the base and acid formulas for each:
Mg(OH)2(aq) + HCl(aq)
?
Al(OH)3(aq) + HCl(aq)
?
STEP 2 Balance OH- and H+ in each:
Mg(OH)2(aq) + 2HCl(aq)
?
Al(OH)3(aq) + 3HCl(aq)
?
General, Organic, and Biological Chemistry
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Solution (continued)
STEP 3 Balance each with H2O:
Mg(OH)2(aq) + 2HCl(aq)
Al(OH)3(aq) + 3HCl(aq)
salt + 2H2O(l)
salt + 3H2O(l)
STEP 4 Write the salt from remaining ions for each:
Mg(OH)2(aq) + 2HCl(aq)
MgCl2(aq) + 2H2O(l)
Al(OH)3(aq) + 3HCl(aq)
AlCl3(aq) + 3H2O(l)
General, Organic, and Biological Chemistry
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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
General, Organic, and Biological Chemistry
Base
(NaOH)
Acid
solution
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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
General, Organic, and Biological Chemistry
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End Point of Titration
At the end point,
 the indicator has a
permanent 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
General, Organic, and Biological Chemistry
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Guide to Calculating Molarity
General, Organic, and Biological Chemistry
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Calculating Molarity from a
Titration with a Base
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 (0.0185 L) of 0.225 M NaOH
10.0 mL of NaOH
Need: M of HCl
STEP 2 Plan:
L of NaOH
moles of NaOH
moles of HCl
M of HCl
General, Organic, and Biological Chemistry
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Calculating Molarity from a
Titration with a Base (continued)
STEP 3 State equalities and conversion factors:
1 L of NaOH = 0.225 mole of NaOH
1 L of NaOH
and 0.225 mole NaOH
0.225 mole NaOH
1 L of NaOH
1 mole of NaOH = 1 mole of HCl
1 mole of NaOH and 1 mole HCl
1 mole HCl
1 mole of NaOH
STEP 4 Set up the problem to calculate moles of HCl:
0.0185 L NaOH x 0.225 mole NaOH x 1 mole HCl
1 L NaOH
1 mole NaOH
= 0.00416 mole of HCl
General, Organic, and Biological Chemistry
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Calculating Molarity from a
Titration with a Base (continued)
STEP 4 (continued)
Calculate the volume in liters of HCl:
10.0 mL HCl = 0.0100 L HCl
Calculate the molarity of HCl:
0.00416 mole HCl = 0.416 M HCl
0.0100 L HCl
General, Organic, and Biological Chemistry
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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
General, Organic, and Biological Chemistry
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Solution
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
STEP 1 Given: 50.0 mL (0.0500 L) of 1.00 M KOH
2.0 M H2SO4
Need: M of H2SO4
STEP 2 Plan:
L of KOH
moles of KOH
moles of H2SO4
mL of H2SO4
General, Organic, and Biological Chemistry
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Solution (continued)
STEP 3 State equalities and conversion factors:
1 L of KOH = 1.00 mole of KOH
1 L of KOH and 1.00 mole KOH
1.00 mole KOH
1 L of KOH
2 moles of KOH = 1 mole of H2SO4
2 moles KOH and 1 mole H2SO4
1 mole H2SO4
2 moles KOH
1 L of H2SO4 = 1000 mL of H2SO4
1 L H2SO4
and 1000 mL H2SO4
1000 mL H2SO4
1 L H2SO4
General, Organic, and Biological Chemistry
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Solution (continued)
1 L of H2SO4 = 2.00 moles of H2SO4
1 L H2SO4
and 2.00 moles H2SO4
2.00 moles H2SO4
1 L H2SO4
STEP 4 Set up the problem to calculate the milliliters
of 2.00 M H2SO4:
0.0500 L x 1.00 mole KOH x 1 mole H2SO4 x
1L
2 moles KOH
1L
x 1000 mL = 12.5 mL
2.00 moles H2SO4
1L
General, Organic, and Biological Chemistry
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