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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) + NaHCO3(s) CO2(g) + NaCl (aq) + H2O(l)

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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) ZnCl2(aq) + H2(g)

B. MgCO3(s) + 2 HCl(aq) MgCl2(aq) + CO2(g) + H2O(l)

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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

Neutralization Reactions

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In the equation for neutralization, an acid and a base produce a salt and water.

acid base salt water

HCl + NaOH NaCl + H2O

2HCl + Ca(OH)2 CaCl2 + 2H2O

Neutralization Equations

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Balancing Neutralization Reactions

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Write the balanced equation for the neutralization ofmagnesium 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

Balancing Neutralization Reactions

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Select the correct group of coefficients for each of thefollowing 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

Learning Check

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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)

Solution

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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 acidHCl and Mylanta.

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Solution

Write the neutralization reactions for stomach acidHCl 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

solutionCopyright © 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 a0.225 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 moles NaOH

x 1 mole HCl = 0.00416 mole HCl 1 mole NaOH

MHCl = 0.00416 mole HCl = 0.416 M HCl

0.0100 L HCl

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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 mL3) 200. mL

Learning Check

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Solution

1) 12.5 mL

0.0500 L KOH x 1.00 mole KOH x 1 mole H2SO4 x

1 L KOH 2 mole KOH

1 L H2SO4 x 1000 mL = 12.5 mL

2.00 mole H2SO4 1 L H2SO4

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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 M2) 0.824 M3) 0.185 M

Learning Check

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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