If a #0.3*mol# quantity of calcium chloride is dissolved in a #1*L# water solvent what are the concentrations of #Ca^(2+)# and #Cl^-#?

2 Answers
Sep 30, 2017

#"Molar concentration"="Moles of solute"/"Volume of solution"#

Explanation:

Now calcium chloride speciates in aqueous solution according to the following chemical equation.....

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

And thus if #0.3*mol# of calcium chloride are dissolved, we gets #0.6*mol# of CHLORIDE ions in solution, and #0.3*mol# of calcium ions....

Capisce?

Sep 30, 2017

0.6

Explanation:

#"Molar concentration"="Number of moles"/"Volume"#

In your example:
#"Number of moles"=0.3 " mol"#
#"Volume"=1000 " ml"#

To change the volume into Liters, we have to divide the volume by thousand (1L contains 1000 ml):
#"Volume"=(1000 " ml")/ (1000 " ml/L") =1" L"#

#rarr "Molar concentration"=0.3/1=0.3 " mol/L"#

Since we are finding the molar concentration of chloride ions:
#CaCl_2rarrCa^(2+)+2Cl^-#

Therefore the concentration of Chloride ions is double the concentration of Calcium chloride:
#"Molar concentration of Chloride"= 2*0.3=0.6" mol/L"#