PLUUUUUls and uncontrollable side reactions. Bromination Ong Ninner The proposed

Question

PLUUUUUls and uncontrollable side reactions. Bromination Ong Ninner The proposed series of equilibria involved generate the electrophilic bromine the reaction to in bromination will be discussed in the report's discussion section. the reagents for One of key used the bromination is sodium hypochlorite (NaCIO found in household bleach. Most bleach bottles will state that the NaCIO concentration is about 5-6%. Unfortunately, the Nacio concentration in bleach depletes over time therefore one of in and the tasks this experiment is to determine that concentration. To determine its , all the NaCIO is reduced in an concentrationof acidic (the browniodine )the titrant standard, sodium solution of iodide T) to yield distinctively -coloured (12. As thiosulfate (NazS,03), is added, iodine is reduced to r by the thiosulfate ion (Scheme 2). NaCIO + 2Nal + 2CHCO H I + NaCl + HO +2 CHCO Na 2SLO,+ IL (brown) – S40 , + 2r (colourless) Scheme 2. redox for NaCIO concentration The reaction determination. Therefore, the endpoint of the titration has been reached when the analyte solution becomes colourless. References

Explanation / Answer

Solution :-

Balanced reaction equations

NaClO +2NaI + 2CH3COOH --- > I2 + NaCl +H2O + 2CH3COONa

2S2O3^2- + I2 ---- > S4O6^2- + 2I-

Lets first calculate the moles of the S2O3^2- used in the titration

Moles of S2O3^2- = molarity * volume

                                  = 0.100 mol per L * 0.0185 L

                                  = 0.00185 mol S2O3^2-

Now using the mole ratio lets calculate the moles of the NaOCl

(0.00185 mol S2O3^2- * 1 mol I2 / 2 mol S2O3^2-)*(1mol I NaOCl/1 mol I2) = 0.000925 mol NaOCl

Now lets convert moles of NaOCl to its mass

Mass of NaOCl = moles * molar mass

                            = 0.000925 mol * 74.44 g per mol

                          = 0.0688 g

So the mass of NaOCl is 0.0688 g

Now lets find the percent of the NaOCl in the bleach

% NaOCl = (mass of NaOCl/ volume of bleach)*100%

                = (0.0688 g / 1 ml )*100%

                = 6.88 %

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