Question #fb741

1 Answer
Mar 15, 2016

Answer:

The first part of the question allows you to determine #K_(sp)#, and the second allows you to use #K_(sp)# to find the solubility of #CaCl_2# in the presence of a common ion (#Cl^-#).

Explanation:

The balanced reaction for ionization of #CaCl_2# is

#CaCl_2(s) harr Ca^(2+)(aq) + 2 Cl^(-)(aq)#
#K_(sp) = [Ca^(2+)][Cl^-]^2#

If #[Cl^-]=2 times 10^-6#M then #[Ca^(2+)] = 1 times 10^-6#M and
#K_(sp) = 4 times 10^-18#

For the second part, the #BaCl_2# produces #[Cl^-]=10^(-2)#M
Therefore, #[Ca^(2+)] = K_(sp)/([Cl^-]^2) = 4 times 10^(-14)#M