Question

Calculate the [OH-] of a 0.20 M aqueous 20
solution of NaF. Ka of HF is 7.1 x 104​

Answer 1

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 :

$\tt [OH^-]=\sqrt{\dfrac{Kw}{Ka}\times M }$

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