What is the ph of 0.0050 hf (ka = 6.8 x 10-4)? 2.73 11.70 11.27 2.30

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

b.final

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error