Answer:
=> [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.)
