1. 12-6 Neutralization and Titration
And you

3. When a strong acid (lots of H+) reacts with a strong base (lots of OH-), the reaction is called a neutralization reaction because the products (water and salt) have neutral pH:
HCl + NaOH → H2O + NaCl

4. The reaction of acids with bases to form water and a salt can be used in the common chemical laboratory technique called titration.
When the moles of acid = moles of base (taking into account the mole ratio), neutralization occurs and the equivalence point of the titration is reached.
At this point, an indicator changes color and the chemist knows to stop the reaction.
A simple calculation can provide the concentration of the unknown acid or base.

5. Example #1: 10 mL 0.01 M HCl neutralized 20 mL NaOH;
what is the molarity of the NaOH? (base)

6. Step 4: Calculate 0.02 L x ?Mb = 0.0001 mole base
Step 1: Reaction HCl + NaOH  NaCl + H2O
Step 2: Mole Ratio 1:1 (1 mole acid neutralizes 1 mole base)
Step 3: Calculate VaMa = Vb Mb
0.01 L acid x 0.01 mole/L acid = mole acid
= requires mole base to neutralize
Step 4: Calculate 0.02 L x ?Mb = mole base
? = M base used

7. I KNOW
It looks complicated – but if you practice I assure you it will not be so daunting!

8. Example #2: (just for you!!!)
80 mL M NaOH neutralized 0.01 M H2SO4. How many mL of acid were present?

9. Step 1: Reaction
H2SO4 + 2NaOH  Na2SO H2O
Step 2: Mole Ratio
1 mole acid neutralizes 2 moles of base
Step 3: Calculate moles of base used
0.08 Lb x mol/Lb = mol base
= mol acid in that sample
Step 4: Calculate ml acid
?mla x 0.01 mol/La = mol acid ?
= 0.01 L ( = 10 mL )

10. Or look at it this way: H2SO4 + 2NaOH  Na2SO4 + 2H2O Base acid
1(vol. base)(molarity base) = 2(Vol. acid)(Molarity acid)
Base acid
1(.08)(.0025) = (?)(.01) = L
note that the acid coefficient is applied to the base side of the equivalency equation – and vice versa.

11. But Why? You may ask

12. 1 mol H2SO4 requires 2 mol NaOH
H2SO NaOH  Na2SO4 + 2H2O
  1 mol H2SO requires mol NaOH
1L of 1M acid requires L of 2M base (or 2L of 1M base)
But if I multiply this relationship by the stoich. Relationships:
1 ( 1L of 1M acid) = (1L of 2M base)
You end up with 1 = 4 and that is just not right!!!!
So switch the stoich. Coefficients and get:
2 ( 1L of 1M acid) = (1L of 2M base)
= much better!!!