Question

Question
Historically chlorine gaz, cl2 has been used do disinfect drinking and swimming pool waters. Knowing that chlorine gas reacts in water according to the following reaction:
CL2+H2O=HOCL+H+CL
and that the manufacturer of bleach water certifies that in 100 g of their product there are 10.8 g of NaCl (the other 89.2 g filling material with no relation at all with disinfection), that HOCl is a weak acid that dissociates as follows:
HOCl = H + OCl , pK= 7.54
Show that the 10.8 g of NaOCl is equivalent to 10.3 g of Cl2

Answer 1

Answer on Question #52337 – Chemistry – Inorganic Chemistry

Question:

Historically chlorine gas, $\mathrm{Cl}_2$ has been used to disinfect drinking and swimming pool waters. Knowing that chlorine gas reacts in water according to the following reaction:

$$\mathrm{CL}_2 + \mathrm{H}_2\mathrm{O} = \mathrm{HOCl} + \mathrm{H} + \mathrm{Cl}$$

and that the manufacturer of bleach water certifies that in $100\,\mathrm{g}$ of their product there are $10.8\,\mathrm{g}$ of NaOCl (the other $89.2\,\mathrm{g}$ filling material with no relation at all with disinfection), that HOCl is a weak acid that dissociates as follows:

$$\mathrm{HOCl} = \mathrm{H} + \mathrm{OCl}, \quad \mathrm{pK} = 7.54$$

Show that the $10.8\,\mathrm{g}$ of NaOCl is equivalent to $10.3\,\mathrm{g}$ of $\mathrm{Cl}_2$.

Answer:

Taking into account the reaction: $\mathrm{CL}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HOCl} + \mathrm{HCl}$

$\mu(\mathrm{Cl}_2) = \mu(\mathrm{HOCl}) = \mu(\mathrm{HCl})$, where $\mu$ - the number of moles.

The formation of NaOCl: $\mathrm{NaOH} + \mathrm{HOCl} + \mathrm{HCl} \rightarrow \mathrm{NaClO} + \mathrm{NaCl} + 2\mathrm{H}_2\mathrm{O}$

Thus, $\mu(\mathrm{Cl}_2)$ must be equal to $\mu(\mathrm{HOCl}) = \mu(\mathrm{HCl}) = \mu(\mathrm{NaCl}) = \mu(\mathrm{NaOCl})$

Calculation of the numbers of moles for NaOCl and $\mathrm{Cl}_2$ leads to:

$$\mu(\mathrm{NaOCl}) = \mathrm{m}/\mathrm{M}_\mathrm{w} = 10.8\,\mathrm{g} / (23 + 16 + 35.5\,\mathrm{g/mol}) = 10.8 / 74.5\,\mathrm{mol} = 0.1450\,\mathrm{mol}$$ $$\mu(\mathrm{Cl}_2) = \mathrm{m}/\mathrm{M}_\mathrm{w} = 10.3\,\mathrm{g} / 71\,\mathrm{g/mol} = 0.1450\,\mathrm{mol}$$

It confirms that $10.8\,\mathrm{g}$ of NaOCl is equivalent to $10.3\,\mathrm{g}$ of $\mathrm{Cl}_2$

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