The active ingredient in many commercial liquid bleaches is sodium hypochlorite. The bottle lists the percentage of sodium hypochlorite as 6.0%. If the density of commercial bleach is 1.084 g/mL, how many mL of 0.150 M sodium thiosulfate is required to reach the end point in a titration similar to the one performed in this experiment, if a student analyzed a 2.0 mL sample of bleach.

23.3mL of 0.150M sodium thiosulfate

Explanation:

The  net reaction of sodium hypochlorite (NaClO) with sodium thiosulfate (Na₂S₂O₃) is:

NaClO + 2 Na₂S₂O₃ + 2H₃O⁺ → NaCl + 3 H₂O  +  Na₂S₄O₆ + 2 Na⁺

2.0mL of the sample of bleach are:

2.0mL ₓ (1.084g / mL) ₓ (6 / 100) ₓ (1 mol / 74.44g) = 1.75x10⁻³ moles of NaClO

As 1 mole of NaClO reacts with 2 moles of thiosulfate:

1.75x10⁻³ moles of NaClO ₓ (2 mol Na₂S₂O₃ / 1 mol NaClO) =

3.50x10⁻³ moles of Na₂S₂O₃

If you have a 0.150M solution of thiosulfate, mL of this solution you need to reach the end point of the titration are:

3.50x10⁻³ moles of Na₂S₂O₃ ₓ (1L / 0.150mol) = 0.0233L =

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.