1. Chapter 8 Acids and Bases
8.5
Reactions of Acids and Bases

2. 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) HCl(aq)  2KCl(aq) + H2(g)
metal acid salt hydrogen gas
Zn(s) HCl(aq)  ZnCl2(aq) + H2(g)
metal acid salt hydrogen gas

3. Learning Check
Write a balanced equation for the reaction of magnesium metal with HCl(aq). Label the metal, acid, and salt.

4. Learning Check
Write a balanced equation for the reaction of magnesium metal with HCl(aq).
Mg(s) HCl(aq)  MgCl2(aq) + H2(g)
metal acid salt

5. 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)
acid carbonate carbon salt water
dioxide
HCl(aq) + NaHCO3(s)  CO2(g) + NaCl (aq) + H2O(l)
acid bicarbonate carbon salt water

6. Learning Check Write a balanced equation for the following reactions:
A. MgCO3(s) + HBr(aq) 
B. HCl(aq) + NaHCO3(aq) 

7. Solution 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

8. Neutralization Reactions
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

9. Neutralization Reactions
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)

10. Guide for Balancing Neutralization Reactions

11. Balancing Neutralization Reactions
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) HNO3
Step 3 Balance the H2O with H+ and the OH.
Mg(OH) HNO3  salt + 2H2O
Step 4 Write the salt from the remaining ions.
Mg(OH) HNO3  Mg(NO3) H2O

12. Learning Check
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, B. 3, 1, 1, C. 3, 1, 1, 3
2. Ba(OH)2(aq) + H3PO4(aq)  Ba3(PO4)2(s) + H2O(l)
A. 3, 2, 2, B. 3, 2, 1, C. 2, 3, 1, 6

13. Solution 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.

14. Solution 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.

15. Basic Compounds in Antacids

16. Learning Check
Write the neutralization reactions for stomach acid, HCl, and the ingredients in Mylanta.
Mylanta: Al(OH)3 and Mg(OH)2

17. Solution 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)

18. Solution 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)

19. 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

20. Indicator An indicator is added to the acid in the flask
causes the solution to change color when the acid is neutralized

21. Endpoint of Titration 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

22. Guide to Calculations for Acid–Base Titrations

23. Acid–Base Titration Calculations
What is the molarity of an HCl solution if 18.5 mL of 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 M NaOH
10.0 mL of HCl
Needed: Molarity of HCl

24. Acid–Base Titration Calculations
What is the molarity of an HCl solution if 18.5 mL of 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 equation molarity
NaOH coefficients HCl
18.5 mL  L  moles NaOH  moles HCl  L HCl

25. Acid–Base Titration Calculations
What is the molarity of an HCl solution if 18.5 mL of 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 and mole HCl x
1 L NaOH mole NaOH

26. Acid–Base Titration Calculations
What is the molarity of an HCl solution if 18.5 mL of 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 mL NaOH  1 L NaOH  mole NaOH
1000 mL NaOH 1 L NaOH
 1 mole HCl = mole of HCl
1 mole NaOH
MHCl = mole HCl = M HCl
L HCl

27. Learning Check
Calculate the mL of 2.00 M H2SO4 required to neutralize mL of 1.00 M KOH.
H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l)
A mL
B mL
C mL

28. Solution
Calculate the mL of 2.00 M H2SO4 required to neutralize mL of 1.00 M KOH.
H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l)
Step 1 State given and needed quantities.
Given: M H2SO4
50.0 mL of 1.00 M KOH
Needed: mL of H2SO4

29. Solution
Calculate the mL of 2.00 M H2SO4 required to neutralize 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 equation molarity
KOH coefficients H2SO4
50.0 mL  L  moles KOH  moles H2SO4  L H2SO4

30. Solution
Calculate the mL of 2.00 M H2SO4 required to neutralize 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 H2SO and mole H2SO4
2.00 mole H2SO mole KOH

31. Solution
Calculate the mL of 2.00 M H2SO4 required to neutralize 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 = L
103 mL
L KOH  1.00 mole KOH  1 mole H2SO4
1 L KOH mole KOH
 1 L H2SO  mL = 12.5 mL.
2.00 mole H2SO L H2SO4
The answer is A.