The [OH-] of NaF solution : 1.7 x 10⁻⁶
Further explanation
Given
0.2 M NaF
Ka of HF is 7.1 x 10⁻⁴
Required
[OH⁻]
Solution
Hydrolysis of NaF salt which is formed from weak acid HF and strong base NaOH
F⁻(aq) + H₂O(l) ⇄ HF(aq) + OH⁻(aq)
General formula :
[tex]\tt [OH^-]=\sqrt{\dfrac{Kw}{Ka}\times M }[/tex]
Input the value :
[OH⁻]=√(10⁻¹⁴/7.1 x 10⁻⁴) x 0.2
[OH⁻]= 1.678 x 10⁻⁶ ≈ 1.7 x 10⁻⁶
