Question

What is the pH in a 1.5 M solution of hydrofluoric acid?

Answer 1

`HF \rightleftharpoons H^(+) + F^(-)\\`

We have 1.5 M HF in the beginning and 0 M
`H^(+)` and
`F^(-)`

We know that there's gonna be a decrease of x M on
`HF` and an increase of x M on
`H^(+)` and
`F^(-)`.

We also find in chemistry tables that
`K_(a)(HF) = 6.8 \cdot 10^(-4)`

`K_a = ([H^+] \cdot [F^-])/([HF])\\6.8 \cdot 10^(-4) = (x \cdot x)/(1.5 - x) \\6.8 \cdot 10^(-4) = (x^2)/(1.5)\\6.8 \cdot 10^(-4) \cdot 1.5 = x^2\\x^2 = 1.02 \cdot 10^(-3)\\x = 0.0319\\pH = -\log{0.0319}\\pH = 1.5`