1. Neutralization Rxns

2. Neutralization Rxns: Please Write Acid + Base  Salt + water
HCl + NaOH  NaCl + H2O
Acid
Base
Salt
H2SO4 + 2KOH  K2SO4 + 2H2O
Acid
Base
Salt

3. Complete the neutralization rxns below:
Please Write
If the acid & base are mixed with equal [H+] and [OH-] a neutral solution will result (pH=7)
Complete the neutralization rxns below:
HNO3 + KOH 
KNO3 + H2O 2
HCl + Mg(OH)2 
MgCl2 + H2O 2
H2SO4 + NH4OH  2
(NH4)2SO4 + H2O 2

4. Acid-Base Titration: Please Write
Is an experimental method that uses the known [acid] to find the [base]. (or reverse)
Base is added to the acid until the solution is neutralized
phenolphthalein will turn pink at a pH > 7
“End point” or “Equivalence point”
At the colour change [H+] = [OH-], for phenolphthalein.
If we measured how much base was needed to neutralize the acid solution, we will know the # OH- moles used & # H+ moles present.

5. How many moles of HCl are required to neutralize 0.20 moles of NaOH?
HCl + NaOH  NaCl + H2O
1 : 1 : 1 : 1
0.20 moles
? : 0.20 moles : :

6. H2SO4 + NaOH  2 Na2SO4 + H2O 2 1 : 2 : 1 : 2 0.25 moles
How many moles of H2SO4 are required to neutralize 0.50 mol of NaOH?
H2SO NaOH  2
Na2SO4 + H2O 2
1 : 2 : 1 : 2
0.25 moles
? : 0.50 moles :

7. 0.15 𝑚𝑜𝑙𝑒𝑠 𝑚𝐿 = 𝑥 500 𝑚𝐿
37.5 mL NaOH used
𝑥= moles of H+
Contains moles of OH-
𝑥 𝑚𝑜𝑙𝑒𝑠 𝑚𝐿 = 𝑚𝑜𝑙𝑒𝑠 37.5 𝑚𝐿
500 mL of 0.15 M HCL
𝑥=2.0 M NaOH

8. Ex. 1 (MH+)(Vacid)=(MOH-)(Vbase)
Please Write
Ex. 1 (MH+)(Vacid)=(MOH-)(Vbase)
MH+ = Macid
monoprotic acids only!
(0.15)(500) =[OH-](37.5) [OH-] = 2.0 M NaOH
Note:
If 0.5M NaOH is used. NaOH  Na OH-
[OH-] = 0.5 M
If 0.5M H2SO4 is used H2SO4  2H SO42-
[H+] = 0.5 x 2 = 1 M

9. (MH+)(Vacid)=(MOH-)(Vbase)
Please Write
Ex.2 Find the volume of 0.25M CH3COOH required to neutralize 35.0 mL of 0.65M KOH solution.
CH3COOH KOH  KCH3COO + H2O
(MH+)(Vacid)=(MOH-)(Vbase)
(0.25)(Vacid) = (0.65)(35.0)
Vacid = 91 mL

10. (MH+)(Vacid)=(MOH-)(Vbase)
Please Write
Ex.3 If 47.3 mL of 1.0M NaOH is required to titrate 25.0 mL of H2SO4 to the equivalence point, determine the concentration of the H2SO4 solution.
H2SO4 + 2NaOH  Na2SO4 + 2H2O
(MH+)(Vacid)=(MOH-)(Vbase)
H2SO4  2 H SO42-
[H2SO4] = [ 𝐻 + ] 2
= [H2SO4] = 0.95M
(MH+)(25.0) = (1.0 M)(47.3)
[H+] = 1.892