Question

What are the total number of ions in solution if a `25.0*g` mass of calcium chloride is dissolved in a `500*mL` volume of water?

Answer 1

`"Concentration"` `=` `"Moles of solute"/"Volume of solution"`

Explanation:

And so we calculate the molar quantity WITH RESPECT to `CaCl_2`, and take the quotient.......

`((25.0*g)/(110.98*g*mol^-1))/(0.50*L)=(0.225*mol)/(0.50*L)=0.450*mol*L^-1`, with respect to `CaCl_2(aq)`

But we know that calcium chloride speciates in solution to give THREE equivalents of ions......

`CaCl_2(s) stackrel(H_2O)rarrCa^(2+) + 2Cl^-`

And thus `[Ca^(2+)]=0.450*mol*L^-1`....and by stoichiometry....

`[Cl^-]=2xx[Ca^(2+)]=0.900*mol*L^-1`

Are you with me? The stoichiometry of the formula determines the respective concentration of the ions in solution. How would this change if we had for instance, `[NaCl(aq)]=0.450*mol*L^-1`; what are `[Na^+]` and `[Cl^-]`?