1. Chapter 4: Concentrations and Titrations

2. Solution Stoichiometry
The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.
M = molarity =
moles of solute
liters of solution
What mass of KI is required to make 500. mL of
a 2.80 M KI solution?
4.5

3. 4.5

4. Acid/Base Titrations
Experimental technique that determines the concentration (in Molarity) of an acid (or base)
This is based upon an acid/base neutralization reaction.
ACID +BASE -> SALT + H2O
Base (or acid) is added until there is the same amount (same # moles) of base and acid.

5. Equivalence point – the point at which the reaction is complete
Titrations
In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.
Equivalence point – the point at which the reaction is complete
Indicator – substance that changes color at (or near) the
equivalence point
Slowly add base
to unknown acid
UNTIL
the indicator
changes color
4.7

6. Fig. 4.17a,b

7. Acid-Base Titrations Base; (OH)- Acid + Base -> Salt + H2O Acid; H+
Introductory Chemistry 2/e by N Tro,
Prentice Hall, 2006, pg 480

8. At the endpoint of an acid/base titration….
Moles acid = Moles base
(MV)acid = (MV)base
Note
If solid; moles = mass/ MM
If aqueous solution; moles = MV

9. What volume of a 1.420 M NaOH solution is
Required to titrate mL of a 4.50 M H2SO4
solution?
WRITE THE CHEMICAL EQUATION!
H2SO4 + 2NaOH H2O + Na2SO4 M
acid rx
coef. M
base
volume acid
moles acid
moles base
volume base
4.50 mol H2SO4
1000 mL soln x
2 mol NaOH
1 mol H2SO4
1000 ml soln
1.420 mol NaOH
25.00 mL
= 158 mL
4.7

10. 1. calculate the number of moles of H2SO4:
How many GRAMS of a M NaOH solution are needed to neutralize 20.0 mL of a M H2SO4 solution?
1. calculate the number of moles of H2SO4:
M = mol/L
0.245 M = mol/ L
mol = 4.90 E -3 mol
2. balance equation and convert to moles of NaOH
2 NaOH + H2SO4  2 H2O + Na2SO4
1 mol H2SO4 = 2 mol NaOH
2 x 4.90 E -3 = 9.80 x 10-3 mol
3. convert to grams of NaOH
9.80 E -3 mol x 40g = g

11. Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.
Dilution
Add Solvent
Moles of solute
before dilution (i)
after dilution (f) =
MiVi
MfVf =
4.5

12. How would you prepare 60.0 mL of 0.2 M
HNO3 from a stock solution of 4.00 M HNO3?
MiVi = MfVf
Vi =
MfVf Mi =
0.200 x 0.06
4.00
= L = 3 mL
3 mL of acid
+ 57 mL of water
= 60 mL of solution
4.5