# What is the pH of a 0.200*mol*L^-1 solution of HF, for which K_a=6.3xx10^-4?

Mar 23, 2017

We interrogate the equilibrium...........

#### Explanation:

We interrogate the equilibrium...........

$H F \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + {F}^{-}$

Unlike the other hydrogen halides, $H F$ is NOT a strong Bronsted acid, and I believe this is an entropy effect. Anyway, we need a ${K}_{a}$ value for the above equilibrium.

This site reports that ${K}_{a}$ for $H F$ $=$ $6.3 \times {10}^{-} 4$. (This really should have been quoted in the question!)

So we set up the equilibrium expression as usual:

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \left[{F}^{-}\right]}{\left[H F\right]} = 6.3 \times {10}^{-} 4$.

Now initially, $\left[H F\right] = 0.200 \cdot m o l \cdot {L}^{-} 1$; we posit that $\text{x} \cdot m o l \cdot {L}^{-} 1$ dissociates, and we substitute these values into the equilibrium expression.

${K}_{a} = \left(\text{x"xx"x")/(0.200-"x}\right) = {x}^{2} / \left(0.200 - x\right) = 6.3 \times {10}^{-} 4$.

This is a quadratic in $x$, which we could solve EXACTLY. Because we are lazy we ASSUME that $0.200 \text{>>} x$, and that $\left(0.200 - x\right) \cong 0.200$. We must justify this approximation later.

So ${x}_{1} = \sqrt{6.3 \times {10}^{-} 4 \times 0.200} = 0.0112 \cdot m o l \cdot {L}^{-} 1$.

Now that we have an approx. value, we can resubstitute this value into the equilibrium expression, and see how our approx. evolves.

${x}_{2} = \sqrt{6.3 \times {10}^{-} 4 \times \left(0.200 - 0.0112\right)} = 0.0109 \cdot m o l \cdot {L}^{-} 1$

${x}_{3} = \sqrt{6.3 \times {10}^{-} 4 \times \left(0.200 - 0.0109\right)} = 0.0109 \cdot m o l \cdot {L}^{-} 1$

Given that our approximations have CONVERGED, we accept $x = 0.0109 \cdot m o l \cdot {L}^{-} 1$.

Now $x = \left[{H}_{3} {O}^{+}\right]$, and thus $\left[{H}_{3} {O}^{+}\right] = 0.0109 \cdot m o l \cdot {L}^{-} 1$, and $p H = 1.96$.