Ch11#1 Calculations Involving Molarity (Titrations).ppt

Calculating that involves molarity, ppt about titrations.

1. Calculations Involving Molarity (Aqueous Solution)
If 100 mL of 1.00 MNaOH and 100 mL of 0.500 MH
2SO
4solution are
mixed, what will the concentration of the resulting salt solution be?
1) Write the balanced equations:
2NaOH + H
2SO
4Na
2SO
4+ 2H
2O
2) Determine the starting # of moles of each reactant:
moles NaOH = 1.00 mol/L * 0.100L = 0.100 mol NaOH
moles H
2SO
4 = 0.500 mol/L * 0.100L =0.0500 mol H
2SO
4
3) Determine the limiting reagant:
moles of Na
2SO
4 from NaOH =
0.100 mol NaOH*(1 mol Na
2SO
4/2mol NaOH)=0.0500mol Na
2SO
4
moles of Na
2SO
4 from H
2SO
4 =
0.0500 mol H
2SO
4*(1 mol Na
2SO
4/1mol H
2SO
4)=0.0500mol Na
2SO
4
4) Determine the final molarity:
Mof Na
2SO
4= 0.0500mol Na
2SO
4/(0.100L +0.100L)=0.250 MNa
2SO
4

2. If 100 mL of 1.00 MNaOH and 100 mL of 0.500 MH
2SO
4solution are
mixed, what will the concentration of the resulting salt solution be?
2NaOH + H
2SO
4 Na
2SO
4 + 2H
2O
Rxn Ratio: 2 mol 1 mol 1 mol 2 mol
Start: 0.100 mol0.0500 mol 0 mol
Change: -0.100 mol -0.0500 mol +0.0500 mol
Determine the final molarity:
Mof Na
2SO
4= 0.0500mol Na
2SO
4/(0.100L +0.100L)=0.250 MNa
2SO
4

3. mmoles (millimoles)
mmoles = 10
-3
moles & mL = 10
-3
L
Molarity = mmol/mL = 10
-3
mol/10
-3
L = mol/L
If 100 mL of 1.00 MNaOH and 100 mL of 0.500 MH
2SO
4solution are
mixed, what will the concentration of the resulting salt solution be?
2NaOH + H
2SO
4Na
2SO
4+ 2H
2O
mmoles NaOH = 1.00 mol/L * 100mL= 100 mmolNaOH
mmoles H
2SO
4 = 0.500 mol/L * 100mL=50.0 mmol H
2SO
4
mmoles of Na
2SO
4 from NaOH =
100 mmolNaOH*(1 mmol Na
2SO
4/2mmol NaOH)=50.0 mmolNa
2SO
4
mmoles of Na
2SO
4 from H
2SO
4 =
50.0mmolH
2SO
4*(1 mmol Na
2SO
4/1mmol H
2SO
4)=50.0 mmolNa
2SO
4
Mof Na
2SO
4= 50.0mmolNa
2SO
4/(100mL + 100mL)=0.250 MNa
2SO
4

4. Titrations
Titration-the process by which one determines the volume of a
standard solution required to react with a specific amount of another
substance.
Standard solutions -solutions of accurately known concentrations.
Primary Standard-a compound that exists in a known high degree of
purity, doesn’t react with the atmosphere, is soluble in water, has a
high formula weight and reacts according to one invariable reaction.
Secondary Standard -A solution that is standardized by a primary
standard.
Standardization-a process by which one determines the concentration
of a solution by titration with a standard solution.
Equivalence point-the point at which chemically equivalent amounts
of acid and base have reacted (neutralization).
End point-the point at which the indicator changes color. (Slightly past
the equivalence point)

5. Indicator-a highly colored substance that can exist in different forms,
with differnet colors that depend on the concentration of protons in soln.

6. Acid-Base Titration
Common Primary Standard -Potassium Hydrogen Phthalate (KHP) an
acidic salt.
C
6H
4(COOH)
2
C
8H
6O
4
phthalic acid
K C
6H
4(COO)(COOH)
KC
8H
5O
4
KHP + NaOH NaKP + H
2O

7. Sodium carbonate is another common primary standard
What is the molarity of a sulfuric acid solution if 40.0 mL of the
solution neutralizes 0.364 grams of sodium carbonate?
Na
2CO
3 + H
2SO
4Na
2SO
4+ H
2O+ CO
2
? Mof sulfuric acid = mol H
2SO
4/L of soln
We need to find the mol of H
2SO
4
0.364 g NaCO
3*(1 mol/106.0g Na
2CO
3)*(1mol H
2SO
4/1mol Na
2CO
3)
= 0.00343 mol H
2SO
4
MH
2SO
4 = 0.00343 mol H
2SO
4/0.0400L H
2SO
4= 0.00858 MH
2SO
4