Traditional Czech and Slovak bleaching and disinfecting products include a commercially available so — Physical Chemistry Chemistry Question
Determination of active chlorine in commercial products
Traditional Czech and Slovak bleaching and disinfecting products include a commercially available solution called “SAVO”. The strength of products like SAVO is commonly expressed as the content of active (or free) chlorine. The active chlorine value determination is usually carried out by adding an excess of potassium iodide to a sample and titrating the equivalent of released iodine with the standard sodium thiosulfate solution.
Consider a reaction (i) between chlorine and water giving products A and B and a reaction, (ii) between sodium hypochlorite and water giving products A and C. Write the chemical equations for both reactions. Which form of A will be dominant in an alkaline aqueous solution?
Model Answer
(i) Cl2 + H2O → HClO (A) + HCl (B)
(ii) NaClO + H2O → HClO (A) + NaOH (C)
In alkaline aqueous solution, hypochlorite ion (ClO-) will dominate.
Commercial SAVO contains 22.4 g dm−3 of active chlorine. What is the corresponding molar concentration of sodium hypochlorite in SAVO?
Model Answer
c(NaClO) = rho(Cl2) / M(Cl2) = 22.4 g dm-3 / 70.906 g mol-1 = 0.3159 mol dm-3
For the determination of sodium hypochlorite, 10.00 cm3 of SAVO (density 1.070 g cm−3) were transferred into a 250 cm3 volumetric flask, distilled water was added to the mark and the solution was mixed. An aliquot of 10.00 cm3 of the solution was transferred into a titration flask and diluted with approx. 50 cm3 of distilled water. After the addition of 5 g of potassium iodide, the released iodine was titrated by 10.15 cm3 of 0.0503 mol dm−3 standard sodium thiosulfate solution to reach the equivalence point. Calculate the mass percentage of sodium hypochlorite in the original SAVO product.
Model Answer
ClO- + 2 I- + 2 H+ → I2 + H2O + Cl-
I2 + 2 S2O3^2- → 2 I- + S4O6^2-
w(NaClO) = c(S2O3^2-) * V(S2O3^2-) * 1/2 * M(NaClO) * [Vflask / (rho_SAVO * V_SAVO * V_a)]
w(NaClO) = 0.0503 * 0.01015 * 1/2 * 74.44 * [0.250 / (1.070 * 0.010 * 0.010)] = 0.0444, i.e. 4.44 %