Question

The label on a bleach bottle states that it contains 5.25 percent (by mass) sodium hypochlrite (NaOCl). Assume that the bleach is manufactured by adding this chemical to otherwise pure water. Take the density of bleach to be the same as that of pure water, 1.0 g cm-3. Assume that NaOCl dissociates completely.

(a) what is the pH of bleach?

(b) Bleach can be used as an emergency disinfectant-for example, for water after an earthquake. Assume that the target does of free residual chlorine (HOCl + OCl-) to be added is 50 gm m-3. How much bleach (cm3) would you add to 1 L of water to disinfect it?

(c) If too much bleach is added to water, it is possible to dechlorinate the water by adding sulfur dioxide (SO2). The sulfur is oxidized from aqueous sulfonic acid to sulfate (SO4-) , while the chlorine is reduced from the hypochlorite ion (OCl- or HOCl) to chloride ion (Cl-). Write a stoichiometrically balanced reaction that show how this dechlorination reaction proceeds.

(d) of course most people dont have SO2 gas cylinders lying around in their home. An alternative dechlorination agent is ammonia (NH3). The nitrogen in ammonia is oxidized to either N20 or N2. Write separate, stoichiometrically balanced reactions for dechlorination by ammonia for each of these end products.

Answer 1

a) Na⁺ is a very weak acid, so there is no effect on pH from this cation. ClO^-, on the other hand, is a moderately weak base, whose hydrolytic activity has to be considered.

The molar concentration of this anion will be the same as the concentration of NaClO:

The equilibrium it is involved in is:

And it is ruled by a basic constant with a value of 3.39x10^-7.

The ICE table for this case is:

	ClO-	HClO	OH-
initial	0.705M	0	0
change	-x	+x	+x
equilibrium	0.705 M - x	x	x

And the equilibrium constant can be written as:

Which is a quadratic equation that can be solved to yield:

x1 is the only answer with chemical meaning, and it represents the concentration of OH^-. We can then calculate:

b) We need a final concentration of 50 g/m³, which is equal to 0.050 g/L. Our solution is 52.5 g/L, so we can calculate the volume needed using the dilution formula:

c) The reaction is:

d) The reactions are: