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.
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