Does a precipitate of barium fluoride form when 100 mL of 1.0 x 10^-3 M Ba(NO3)2(aq) is mixed with 200 mL of 1.0 x 10^-3 M KF(aq)? Ignore possible protonation of F-.

Each solution is diluted by mixing. The Volume of the mixture is 100 mL + 200 mL = 300 mL.

[Ba2+] in the mixture is 1.0 x 10^-3 M x 100mL/300mL = 0.3x 10^-3 M

[F-] in the mixture is 1.0 x 10^-3 M x 200mL/300mL = 0.67x 10^-3 M

Test for the solubility product:

[Ba2+]x[F-]^2= 0.3x 10^-3 M x 0.67^2x 10^-6 M = 0.13x 10^-9M^3 that is lower than the solubility product of BaF2, Ksp = 1.0x10^-6.

Answer: no precipitate.

Secure Information

Content will be erased after question is completed.

Enter the email address associated with your account, and we will email you a link to reset your password.

Forgot your password?

Sign Up