Question

What is the pH of 0.0050 HF (Ka = 6.8 x 10-4)? 2.73 11.70 11.27 2.30

Answer 1

The correct answer is 2.73.

HF is a weak acid which partially dissociates to release H+ and F-

                HF           →         H⁺   + F⁻

Initial      0.0050                   0       0

Change    -x                        +x       +x

Equilibrium 0.0050–x         +x     +x

Solve by using the equilibrium expression: =  [H⁺] [F⁻]/ [HF]

6 .8 x 10⁻⁴= x. x / 0.0050   –x

6 .8 x 10⁻⁴= x² /0.0050  

x²  = 6 .8 x 10⁻⁴ x 0.0050  

x²  = 3.4 x 10⁻⁶

x = 3.4 x 10⁻⁶

[H⁺]  = 1.84 x 10⁻³

pH = - log [H⁺] = - log (1.84 x 10⁻³)

pH = 2.73