The K_a of hydrofluoric acid, HF, is 6.8 x 10^{-4} - VCE - SSCE Chemistry - Question 17 - 2004 - Paper 1

To find the pH, we first establish the dissociation of HF in water:

$HF \leftrightarrow H^+ + F^-$

Given that the ionization constant is ( K_a = 6.8 \times 10^{-4} ), we can set up the equation:

$K_a = \frac{[H^+][F^-]}{[HF]}$

Assuming ( x ) is the concentration of ( H^+ ) ions produced, we can express:

• Initial concentration of HF = 0.10 M
• Change = -x
• At equilibrium:
  • ( [HF] = 0.10 - x \approx 0.10 ) (since x will be small)
  • ( [H^+] = x )
  • ( [F^-] = x )

Thus, we have:

$6.8 \times 10^{-4} = \frac{x^2}{0.10}$

Rearranging gives:

$x^2 = 6.8 \times 10^{-4} \times 0.10 = 6.8 \times 10^{-5}$

Taking the square root to find ( x ):

$x = \sqrt{6.8 \times 10^{-5}} \approx 0.00824 M$(approximately)

Therefore, ( [H^+] = 0.00824 M $.