A solution contains 3.8 × 10-2 M in Al3+ and 0.29 M in NaF. If the Kf for AlF6 3- is 7 × 1019, how much aluminum ion remains at equilibrium? A) 9.1 × 10-19 M B) 1.1 × 10-19 M C) 4.4 × 10-20 M D) 3.1 × 10-22 M E) 1.9 × 10-21 M
Added by Taylor H.
Step 1
Step 1: Write equilibrium for formation: Al3+ + 6 F- ⇌ AlF6 3- with Kf = [AlF6 3-]/([Al3+][F-]^6) = 7×10^19. Show more…
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