# Question dd490

Nov 22, 2015

1.7

#### Explanation:

After the addition of HF, but before any dissociation can occur, the concentration of HF, ["HF"] = 0.750" M". $\left[{\text{F}}^{-}\right] = 0$ and $\left[{\text{H"_3"O}}^{+}\right]$ is negligible.

We need to know the amount of HF dissociated. Since the volume of the solution, $V$, is assumed to be constant, denote $x$ to be the amount of HF dissociated divided by $V$. This is to facilitate calculations later since the acid dissociation constant is expressed in terms of the product of the concentration of the species involved.

After dissociation,
["HF"] = 0.750" M"-x#
$\left[{\text{F"^-]=["H"_3"O}}^{+}\right] = 0 + x = x$

At equilibrium,

$\frac{\left[\text{F"^-]*["H"_3"O"^+]}{["HF}\right]}{=} {K}_{a}$.

Therefore,

$\frac{{x}^{2}}{0.750 - x} = 6.8 \times {10}^{- 4}$.

Solving the quadratic gives

$x = 0.0222$

Since $\left[{\text{H"_3"O}}^{+}\right] = x$, the pH of the solution is given by

$- \text{lg} \left(0.0222\right) = 1.7$