Calculate the [H+] and pH of a 0.0040 M hydrofluoric acid solution. The Ka of hydrofluoric acid is 6.80 × 10-4. Use the method of successive approximations in your calculations.

Answer 1

HF ==> H⁺ + F^- ... ionization of hydrofluoric acid

Ka = [H+][F-] / [HF]

6.80x10^-4 = (x)(x) / 0.0040 - x

6.80x10^-4 = x² / 4.0x10^-3 - x

Using the method of successive approximations, we will assume x is small and ignore it in the denominator

6.80x10^-4 = x² / 4.0x10^-3

x² = (6.80x10^-4 )(0.0040) = 2.72x10^-6

x = 1.65x10^-3

Substituting this back into the original equation and solving for x, we have ...

6.80x10^-4 = x² / 4.0x10^-3 - 1.65x10^-3 = x² / 2.35x10^-3

x² = 1.6x10^-6

x = 1.3x10^-3

Substituting this back into the original equation and solving for x, we have ...

6.80x10^-4 = x² / 4.0x10^-3 - 1.3x10^-3 = x² / 2.7x10^-3

x² = 1.8x10^-6

x = 1.35x10^-3 (same as previous iteration)

[H⁺] = 1.35x10^-3 M

pH = -log [H⁺]

pH = -log 1.3x10^-3

pH = 2.9

(be sure to check all of the math)