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)

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- + 2 I-  I2 + Cl- + H2O 2) 3) I2 + I-  I3- 4) I3- + 2 S2O32-  3 I- + S4O62-

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

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- + 2 I-  I2 + Cl- + H2O 2) 3) I2 + I-  I3- The titration: 4) I3- + 2 S2O32-  3 I- + S4O62-

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

11.3 ml of .100 M Na2S2O3 X moles Na2S2O3 .100 M Na2S2O3 = .0113 L Na2S2O3 .00113 moles Na2S2O3 = .00113 moles S2O3- 1 mole ClO- .00113 moles S2O3- x 2 moles S2O3- = .000565 moles ClO-

of the diluted bleach that was titrated! = .000565 moles ClO- = .000565 moles NaClO that were in a 25 ml sample of the diluted bleach that was titrated! .000565 moles NaClO M = .025 L M = .0226 M diluted bleach

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

% 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 %