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

The [OH-] of NaF solution : 1.7 x 10⁻⁶

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 :

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

the mantle

explanation:

earth is divided into three layers: the crust or the outermost layer on which we live; the underlying, middle layer - the mantle - and the core (the innermost layer of earth).

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i