1. Analysis of Bleach NaClO - sodium hypochlorite
Na+ and ClO- hypochlorite ion
Purpose: To determine the concentration of the bleach (NaClO)
Molarity and % by mass (percent composition)

2. Equations 1) 2 H+ + ClO- + 2 I-  I2 + Cl- + H2O 2) 3) I2 + I-  I3-
Cl2 + NaOH(aq)  ClO- + Cl- H2O + Na+
Cl2 + Na+ + OH-  ClO- + Cl- H2O + Na+
Cl2 + 2OH-  ClO- + Cl- + H2O 1)
2 H+ + ClO I-  I2 + Cl- + H2O 2) 3)
I I-  I3- 4)
I S2O32-  3 I- + S4O62-

3. 2 A + 1 B  3 C + 2 D 1 C + 3 E  4 F 2 F  3 G What is the ratio of…
- A’s to C’s?
- A’s to E’s?
- A’s to F’s?
- A’s to G’s?
2 A’s : 3 C’s
2 A’s : 9 E’s
1 A’s : 6 F’s
1 A’s : 9 F’s

4. The titration: 2 H+ + ClO- + 2 I-  I2 + Cl- + H2O 2) 3) I2 + I-  I3-
from your diluted
bleach solution
from the
HCl solution
from the
KI crystals
2 H+ + ClO I-  I2 + Cl- + H2O 2) 3)
I I-  I3-
The titration: 4)
I S2O32-  3 I- + S4O62-

5. DARK BLUE! The titration: 4) I3- + 2 S2O32-  3 I- + S4O62-
yellow
orange
brownish red
colorless
colorless
colorless
inc.
conc.
in starch
in starch
I S2O32-  3 I- + S4O62-
DARK
BLUE!
colorless

6. 11.3 ml of .100 M Na2S2O3
X moles Na2S2O3
.100 M Na2S2O3 =
.0113 L Na2S2O3
moles Na2S2O3
= moles S2O3-
1 mole ClO-
moles S2O3- x
2 moles S2O3-
= moles ClO-

7. of the diluted bleach that was titrated!
= moles ClO-
= moles NaClO
that were in a 25 ml sample
of the diluted bleach that was titrated!
moles NaClO
M =
.025 L
M = M diluted bleach

8. (Mconc. ) (Vconc.) = (Mdil. ) (Vdil.)
M = M diluted bleach
(Mconc. ) (Vconc.) = (Mdil. ) (Vdil.)
(Mconc. ) (5 mlconc.) = (.0226 Mdil. ) (100 mldil.)
Mconc. = .452 M

9. % mass of sodium hypochlorite = 3.12 %
Mconc. = .452 M
.452 mol NaClO
Density = 1.08 g/ml
M =
1 L
.452 mol NaClO
Density = 1080 g/L
1080 g
.452 mol x 74.5 g/mol
33.7 g NaClO =
1080 g
1080 g of sol.
% mass of sodium hypochlorite = 3.12 %