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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. 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.

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chapter 8 acids and bases
Chapter 8 Acids and Bases

8.6

Reactions of Acids and Bases

Copyright © 2005 by Pearson Education, Inc.

Publishing as Benjamin Cummings

acids and metals
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)

acids and carbonates
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)

learning check
Learning Check

Write the products of the following reactions of acids.

A. Zn(s) + 2 HCl(aq)

B. MgCO3(s) + 2HCl(aq)

solution
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)

neutralization reactions
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

neutralization equations
Neutralization Equations

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

acid base saltwater

HCl + NaOH NaCl + H2O

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

balancing neutralization reactions9
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 + 2HNO3Mg(NO3)2 + 2H2O

learning check10
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

solution11
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)

basic compounds in some antacids
Basic Compounds in Some Antacids

Antacids are used to neutralize stomach acid (HCl).

learning check13
Learning Check

Write the neutralization reactions for stomach acid

HCl and Mylanta.

solution14
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)

acid base titration
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

indicator
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

end point of titration
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

calculating molarity
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 1Given: 18.5 mL of 0.225 M NaOH; 10.0 mL HCl

Need: Molarity of HCl

STEP 218.5 mL L moles NaOH moles HCl M HCl

L HCl

STEP 31 L = 1000 mL 0.225 mole NaOH/1 L NaOH

1 mole HCl/1 mole NaOH

calculating molarity continued
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

learning check20
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

solution21
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

learning check22
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

solution23
Solution

2) 0.824 M

0.0426 L x 1.45 mole NaOH x 1 mole H3PO4

1 L3 mole NaOH

= 0.0206 mole H3PO4

0.0206 mole H3PO4 = 0.824 mole/L = 0.824 M

0.0250 L