Answer to Question #303800 in Physical Chemistry for Chemistry

Question

Hydrofluoric acid has a Ka value of 7.2x10−4. What is the pH of a 0.04M solution

of hydrofluoric acid?

Answer 1

$7.2×10^{-4}=\frac{[H^+][F^-]}{[HF]}$

$=\frac{(x)(x)}{(0.04-x)}$

Ignoring $x$ in the denominator

$x^2=2.88×10^{-5}$

$x=5.367×10^{-3}$

pH$=-$ log $(5.367×10^{-3})$

$=2.3$