6. Calculate the [OH-] of a 0.200 M aqueous solution of Naf. Ka of HF is 6.7 x 10-4. (Kw = 1 x 10-14 *
(3 Points)
1.2 x10-5
2.0 x10-6
6.7 c10-4
3.9.x10-6
o 1.7x10-6

=> [OH⁻] = 2.0 x 10⁻⁶M (1 sig. fig.)

Explanation:

Given 0.200M NaF => 0.200M Na⁺(aq) + 0.200M F⁻(aq)

Na⁺ + H₂O => no rxn (sodium ion will not hydrolyze to form weak electrolyte

F⁻ + H₂O => HF + OH⁻ (F⁻ hydrolyzes in water to form weak acid HF)

Analysis:

                  F⁻        +    H₂O   =>     HF     +      OH⁻

C(i)):      0.200M                 0.00M       0.00M

ΔC:          -x                            +x              +x

C(eq): 0.200-x                          x                x

         ~0.200M

Kb = Kw/Ka(HF) = [HF][OH⁻]/[F⁻] = x²/(0.200) = 1x10⁻¹⁴/6.7x10⁻⁴

=> x = [HF] = [OH⁻] = √(0.200)(1x10⁻¹⁴)/(6.7x10⁻⁴) = 1.73 x 10⁻⁶M in OH⁻ ions.

=> [OH⁻] = 2.0 x 10⁻⁶M (1 sig. fig.)

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