The bottle of bleach lists the percentage of sodium hypochlorite as 6.0%. If the density of commercial bleach is 1.084 g/mL, how many mL of .150 M sodium thiosulfate is required to reach the endpoint in a titration if a student analyzed a 2.0 mL sample of bleach.
The first thing I would do is calculate the molarity of the bleach.
That is 1.084 g/mL x 1000 mL x 0.06 = 65.04 g and that divided by the molar mass NaOCl = 65.04/74.44 = 0.874M. Therefore, a 2 mL sample will have mols = M x L = 0.874 x 0.002L = 0.00175 mols.
Are you treating the NaOCl with I^- to oxidize it to I2 then titrating the liberated I2 with thiosulfate?
OCl^- + 2I^- ==>I2 + Cl^-
2S2O3^2- + I2 ==>2I^- + S4O6^2-
1 mol OCl^- = 1 mol I2 = 2 mol S2O3
Therefore 1/2 mol NaOCl = 1 mol S2O3
We have 0.00175 mol NaOCl. That will use 1/2 that of thiosulfate.
M thiosulfate = mols/L
You know M and mols; solve for L and convert to mL.
To find out how many mL of 0.150 M sodium thiosulfate is required to reach the endpoint in a titration, we first need to determine the moles of sodium hypochlorite in the sample of bleach.
1. Begin by converting the percentage of sodium hypochlorite to grams. Since the density is given in g/mL, we can assume that the sample size of bleach is 2.0 mL. Therefore, the mass of sodium hypochlorite in the sample is calculated as follows:
mass = volume × density
= 2.0 mL × 1.084 g/mL
= 2.168 g
2. Next, convert the mass of sodium hypochlorite to moles. The molar mass of sodium hypochlorite (NaOCl) can be calculated using the atomic masses of its constituent elements:
molar mass of NaOCl = atomic mass of Na + atomic mass of O + atomic mass of Cl
= (22.99 g/mol) + (16.00 g/mol) + (35.45 g/mol)
= 74.44 g/mol
moles = mass / molar mass
= 2.168 g / 74.44 g/mol
≈ 0.0291 mol
3. In a titration, sodium thiosulfate (Na2S2O3) reacts with sodium hypochlorite based on their stoichiometric ratio. From the balanced chemical equation:
2 NaOCl + Na2S2O3 → NaCl + Na2SO4 + S
There is a 1:1 stoichiometric ratio between sodium hypochlorite and sodium thiosulfate. This means that for every mole of sodium hypochlorite, we need an equal number of moles of sodium thiosulfate.
So, the moles of sodium thiosulfate required would also be approximately 0.0291 mol.
4. Finally, to calculate the volume of 0.150 M sodium thiosulfate needed, we can use the equation:
moles = concentration × volume
Rearranging the equation, we can solve for volume:
volume = moles / concentration
= 0.0291 mol / 0.150 mol/L
≈ 0.194 mL
Therefore, approximately 0.194 mL of 0.150 M sodium thiosulfate is required to reach the endpoint in the titration when analyzing a 2.0 mL sample of bleach.