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?

1 Answer
Jul 20, 2017

Answer:

#"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^-]#?