Indicate the concentration of each ion present in the solution formed by mixing: 3.40g KCl in 75.0 mL of 0.210 M CaCl2 solution. Assume that the volumes are additive.

Respuesta :

To solve this task, we have to use this molar formula : Number of Moles=Mass/Molar Mass.
According to this, we can calculate each : 
[tex]KCl=3.40/39+35.5[/tex] which is 0,045 moles.
Then use this formula to go on :
№of moles=Concentration(mol/dm3)∗Volume(dm3), and  then [tex]CaCl2=75*0.210/1000 [/tex], which is 0.01575Moles.
Then you can calculate concentration :
[tex]K+=0.045 Cl-=0.045[/tex]
[tex]Ca+2=0.01575 Cl-=0.01575*2[/tex]
Total number : [tex]Cl-=(0.01575*2)+0.045[/tex]
Here are moles of K+ 0.045/0.075
Of Cl- 
(0.01575*2)+0.045/0.075
Ca2+ 
0.01575/0.075
And the needed answer is definitely : 
=0.06075