Question

Suppose 50.0 mL of 0.250 M CoCl2 solution is added to 25.0 mL of 0.350 M NiCl2 solution. Calculate the concentration, in moles per liter, of each of the ions present after mixing. Assume that the volumes are additive.

Answer 1

For CoCl2 the new concentration will be 0.166 M and for the NiCl2 will be 0.116 M

Step-by-step explanation:

When you mix two solutions, the first thing is that the total volume will increase, and to calculate the new concentration that will be smaller you can use the formula: V1C1 = V2C2 in which V1 and C1 are the volume and the concentration at the start point and V2 and C2 are the volume and the concentration at the end of the dilution.

so for the first one,

(50 ml ) (0.250 M of CoCl2) = (75 ml) ( x)

X = (50 x 0.250) / 75 = 0.166 M of CoCl2

for the second one,

(25 ml) (0.350 M NiCl2) = (75 ml) ( x)

x= (25 x 0.350 ) /75 = 0.116 M of NiCl2

Hope this info is useful