Concentration of Stock

Question

the concentration of stock solution X is 3.6 x 10^-3 M. The concentration of stock solution Y is .15M. If 4.0 mL of X is mixed with 6.0 mL of Y , what is the new concentrations of X and Y?

Answer 1

You can use V1M1 = V2M2 for each solution separately.  Volume 2 will be 4.0 ml + 6.0 ml = 10.0 ml.  Thus...

For X:  (4.0 ml)(3.6x10^-3M) = (10.0 ml)(M₂) and M₂ = 1.44x10^-3 M = 1.4x10^-3 M 

For Y:  (6.0 ml)(0.15 M) = (10.0 ml)(M₂) and M₂ = 0.090 M