2007-04-12 17:23:24 · 3 answers · asked by monkeynurse 2 in Science & Mathematics ➔ Chemistry
The Ka equation for HF is [F-][H+]/[HF]=6.8x10-4
Consider a 1 liter solution. HF dissociates slightly, and for each x moles dissociated, x g-ion of H+ and F- are formed. We know that the [H+] conc is sufficient to make the solution pH 3.15.
So [H+] = 10-4 - log 0.85. = 6x10-4 mol/L. Then if we started with M moles of HF,
(6x10-4)^2/(M-x) =6.8x10-4
36x10-8/ 6.8x10-4 = (M-x)
(M-x) = 5.3x10-4 appx
Since x is 6x10-4, M= 11.3x10-4 mol/L
In 0.16 L, we need 11.3x0.16x10-4 mol=
1.8x10-4 moles appx
2007-04-12 17:39:48 · answer #1 · answered by cattbarf 7 · 1⤊ 0⤋
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2016-05-16 12:39:11 · answer #2 · answered by Anonymous · 0⤊ 0⤋
HF Ka = 6.8x10^-4 pH = 3.15 [H+] = 7.08x10^-4M as a results of fact [H+] = [F-] as a results of dissociation, [F-] = 7.08x10^-4M 6.8x10^-4 = (7.08x10^-4)^2 / [HF] [HF] = 7.34x10^-4M 0.15L x 7.34 x 10^-4M = a million.x10^-4moles
2016-12-16 04:26:51 · answer #3 · answered by ? 4 · 0⤊ 0⤋