What is the concentration of chloride ion in a solution composed by mixing 447.2*mL of 0.216*mol*L^-1 NaCl, and 100.8*mL of 0.358*mol*L^-1 CaCl_2. Why would this not be applicable to a solution of PCl_3 added to water?

1 Answer
Sep 15, 2017

This question is very poorly proposed.....

Explanation:

We can assess [Cl^-] with respect to the sodium chloride and the calcium chloride solutions.....We cannot do so in relation to the PCl_3 as we will see later.....

For NaCl and CaCl_2 we assume that the concentrations are ADDITIVE, and we use the old ratio....

"concentration"="moles of solute"/"volume of solution"

And thus "moles of solute"="concentration"xx"volume"

"Moles of NaCl"=447.2xx10^-3*Lxx0.216*mol*L^-1=0.0966*mol" (i)"

"Moles of CaCl"_2=100.8xx10^-3*Lxx0.358*mol*L^-1=0.0361*mol" (ii)"

And so we take....((i)+2xx(ii))/("combined volume")

[Cl^-]=(0.169*mol)/(0.548*L)=0.308*mol*L^-1....

As for the PCl_3, it is a fact that liquid PCl_3 REACTS with water to give phosphorous acid and hydrogen chloride....

PCl_3(l) + 3H_2O(l) rarr H_3PO_3(aq) + 3HCl(aq)

And thus we could not have a solution of PCl_3 in water or in ethanol; we COULD have such a solution in petroleum ether or methylene chloride, but these solvents are insoluble in water.

This reaction between phosphorus trichloride and water is very violent and not one that I would do with these quantities.....and so the question is VERY poorly proposed....