What is the molarity of NaOCL in a solution of bleach if 10.00mL requires 22.35mL of 0.0291 M Na2S2O4 to reach the endpoint.
There must be more to it than that. NaOCl can't be titrated with thiosulfate, at least not when I went to school. The USUAL process is to treat a solution of bleach with KI and titrate the liberated I2 with thiosulfate.
To determine the molarity of NaOCL in a solution of bleach, we need to use a titration calculation. In a titration, a known concentration of one substance (the titrant) is reacted with an unknown concentration of another substance until a chemical reaction reaches its endpoint. In this case, the titrant is a solution of Na2S2O4, and the substance being titrated is NaOCL in bleach.
To calculate the molarity of NaOCL, we need to identify the balanced chemical equation for the reaction between Na2S2O4 and NaOCL. From the equation, we can determine the stoichiometry and the number of moles of Na2S2O4 consumed relative to NaOCL.
The balanced chemical equation for the reaction between Na2S2O4 and NaOCL is:
2 NaOCL + Na2S2O4 + H2O → 2 NaCl + Na2SO4 + 2 H2SO4
From the equation, we can see that it requires 2 moles of NaOCL to react with 1 mole of Na2S2O4. This means that the molar ratio of NaOCL to Na2S2O4 is 2:1.
Next, we can calculate the number of moles of Na2S2O4 used in the titration. The volume of Na2S2O4 used is 22.35 mL, and the molarity of Na2S2O4 is 0.0291 M. Using the formula:
moles = volume (in liters) x molarity
we can convert the volume of Na2S2O4 used to liters:
22.35 mL ÷ 1000 mL/L = 0.02235 L
Now, we can calculate the number of moles of Na2S2O4:
moles of Na2S2O4 = 0.02235 L x 0.0291 M = 0.000649485 moles
Since the molar ratio of NaOCL to Na2S2O4 is 2:1, we can then calculate the number of moles of NaOCL in the original 10.00 mL solution using the same molar ratio:
moles of NaOCL = 2 x 0.000649485 moles = 0.00129897 moles
Finally, we can calculate the molarity of NaOCL in the bleach solution using the formula:
molarity = moles / volume (in liters)
Volume in liters = 10.00 mL ÷ 1000 mL/L = 0.01 L
molarity = 0.00129897 moles / 0.01 L = 0.1299 M
Therefore, the molarity of NaOCL in the bleach solution is 0.1299 M.