Question

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​

Answer 1

=> [OH⁻] = 2.0 x 10⁻⁶M (1 sig. fig.)

Step-by-step 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.)