Transcript Acid-Base Reactions

10.2 Neutralization and Acid-Base
Titrations
Learning Goal …
… use Stoichiometry to calculate volumes and
concentrations in a neutralization reaction
… determine the pH of a solution with an excess
reactant
recall:
Acid
+
Base --->
Water + salt
• this is known as neutralization
• a SALT is an ionic compound that is composed of the anion
from an acid and a cation from a base
HNO3 (aq) + NaOH (aq) 
anion
cation
acid
+ base

H2O (l)
+
water
+
NaNO3 (aq)
salt
• if we mix equal amounts of the acid and the base, all acidic and
basic properties will be neutralized and the pH of the mixture
will be 7
• to neutralize an acid or a base we can thus add equal number of
moles of the opposite substance
What volume of 0.250 mol/L sulfuric acid is needed to
react completely with 37.2 mL of 0.650 mol/L potassium
hydroxide?
H2SO4 (aq) + 2 KOH (aq)  K2SO4 (aq) + 2 H2O (l)
n
0.01209 mol
C
0.250 mol/L
0.650 mol/L
V
0.0484 L
0.0372 L
0.02418 mol
 48.4 mL of sulfuric acid is needed to completely react with the
KOH.
25.0 mL of 1.0 M hydrochloric acid reacts with 25.0 mL
of a 0.50 mol/L solution of NaOH (aq). How many moles
of excess reactant remain? What is the concentration of
excess reactant? What is the pH of the final solution?
NaOH (aq) + HCl (aq)  NaCl(aq) +
n
0.0125 mol
0.025 mol
C
0.50 mol/L
1.0 mol/L
V
0.0250 L
0.0250 L
i
0.0125 mol
0.025 mol
u
0.0125 mol
0.0125 mol
xs
0
0.0125 mol
H2O (l)
[HCl] = n / V
= 0.0125 mol / 0.050 L
= 0.25 mol/L
pH = - log (0.25)
= 0.60
 0.0125 mol of excess HCl remains, the [HCl] is 0.25 mol/L
and the pH of the final solution is 0.60.
CAN I …
… use Stoichiometry to calculate volumes and
concentrations in a neutralization reaction
HOMEWORK
p466 #1-5
Read p466-469 “Titrations”
p470 #1-9