Ba(NO3)^2 is added to a solution of 0.025M of NaF.
-- At what conc. of Ba^2+ does a precipitate start to form?
Here's the start of my work
BaF2 --> Ba^2+ + 2F-
Ksp = [Ba^2+] [F-]^2
what should my next step be?
2007-06-04 13:34:43 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
You need to find the Ksp of BaF2 from whatever table your class uses. And then insert the concentration of the Fluoride (0.025 since NaF dissolves completely) and solve for the concentration of Ba2+
2007-06-04 13:43:16 · answer #1 · answered by Calypso 1 · 0⤊ 0⤋
Your next step: The Ksp of BaF2 = 2.4x10^-5
Let [Ba 2+] = b Then [F-] = 2b + 0.025 ~ 0.025 = 2.5x10^-2 Because b is so small, we can neglect it in the [F-].
(b)(2.5x10^-2)^2 = 2.4x10^-5
b(6.25x10-4) = 2.4x10^-5
b = 2.6x10^-9M
2007-06-04 21:36:13 · answer #2 · answered by steve_geo1 7 · 0⤊ 0⤋