1. Chapter 10 Acids and Bases
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
Equations:
2K(s) + 2HCl(aq) 2KCl(aq) + H2(g)
Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)
Net ionic equations:
2K(s) + 2H+(aq) 2K+(aq) + H2(g)
Zn(s) + 2H+(aq) Zn2+ (aq) + H2(g)

3. 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)

4. Learning Check A. Zn(s) + 2 HCl (aq) ? B. MgCO3 (s) + 2HCl(aq) ?
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) ?

5. 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) MgCl2(aq) + CO2(g) + H2O(l) MgCO3(s) + 2H+(aq) Mg2+(aq) + H2(g)

6. 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) H2O(l)

7. 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) NaCl(aq) + H2O(l)
H+(aq) + OH−(aq) H2O
2HCl(aq) + Ca(OH)2(aq) CaCl2(aq) + 2H2O(l)
H+(aq) + OH−(aq) H2O(l)

8. Guide to Balancing an Equation for Neutralization

9. 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)

10. 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, ) 3, 1, 1, ) 3, 1, 1, 3
B. Ba(OH)2(aq) + H3PO4(aq) Ba3(PO4)2(s) + H2O(l)
1) 3, 2, 2, ) 3, 2, 1, ) 2, 3, 1, 6

11. 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)

12. Antacids
Antacids
are used to neutralize stomach acid (HCl)

13. Learning Check Write the neutralization reactions for stomach acid
HCl and Mylanta.

14. 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) ?

15. Solution (continued) STEP 3 Balance each with H2O:
Mg(OH)2(aq) + 2HCl(aq) salt + 2H2O(l)
Al(OH)3(aq) + 3HCl(aq) 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)

16. 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

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

18. 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

19. Guide to Calculating Molarity

20. 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 ( L) of 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

21. Calculating Molarity from a Titration with a Base (continued)
STEP 3 State equalities and conversion factors:
1 L of NaOH = mole of NaOH
1 L of NaOH and mole NaOH
0.225 mole NaOH L of NaOH
1 mole of NaOH = 1 mole of HCl
1 mole of NaOH and 1 mole HCl
1 mole HCl mole of NaOH
STEP 4 Set up the problem to calculate moles of HCl:
L NaOH x mole NaOH x 1 mole HCl
1 L NaOH mole NaOH
= mole of HCl

22. Calculating Molarity from a Titration with a Base (continued)
STEP 4 (continued)
Calculate the volume in liters of HCl:
10.0 mL HCl = L HCl
Calculate the molarity of HCl:
mole HCl = M HCl
L HCl

23. Learning Check H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l) 1) 12.5 mL
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) mL
2) mL
3) 200 mL

24. 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) mL
STEP 1 Given: 50.0 mL ( 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

25. Solution (continued) STEP 3 State equalities and conversion factors:
1 L of KOH = mole of KOH
1 L of KOH and mole KOH
1.00 mole KOH L of KOH
2 moles of KOH = 1 mole of H2SO4
2 moles KOH and 1 mole H2SO4
1 mole H2SO moles KOH
1 L of H2SO4 = mL of H2SO4
1 L H2SO and mL H2SO4
1000 mL H2SO L H2SO4

26. Solution (continued) 1 L of H2SO4 = 2.00 moles of H2SO4
1 L H2SO and moles H2SO4
2.00 moles H2SO L H2SO4
STEP 4 Set up the problem to calculate the milliliters of 2.00 M H2SO4:
L x 1.00 mole KOH x 1 mole H2SO4 x
1 L moles KOH
1 L x mL = mL
2.00 moles H2SO L