Question

Question #fb741

Answer 1

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