Question

The ionization constant of $HF$ is $3.2\times {10}^{-4}$. Calculate the degree of dissociation of $HF$ in its $0.02M$ solution. Calculate the concentration of all species present $(H_3{O}^{+},F^{-}\text{and}HF)$ in the solution and its pH.

Answer 1

Step 1: degree of dissociation

The following proton transfer reactions are possible:

$1)HF+H_{2}O\rightleftharpoons H_3{O}^{+}+F^{-};K_a=3.2\times {10}^{–4}$

$2)H_{2}O+H_{2}O\rightleftharpoons H_3{O}^{+}+O{H}^{-};K_w=1.0\times {10}^{-14}$

As $K_a>>K_w,\left[1\right]$ is the principal reaction.


$HF+H_{2}O\rightleftharpoons H_3{O}^{+}+F^{-}$

Initial conc. (M) $0.0200$

Equilibrium conc. (M) $0.02(1–\alpha )0.02\alpha 0.02\alpha$


Now equilibrium constant $\left(K_a\right)=\frac{\left[F^{-}\right]\left[H_3{O}^{+}\right]}{\left[HF\right]}$

$K_a=\frac{(0.02\alpha {)}^2}{0.02-0.02\alpha }$

$\Rightarrow \frac{0.02{\alpha }^2}{1-\alpha }=3.2\times {10}^{-4}$

We obtain the following quadratic equation:

${\alpha }^2+1.6\times {10}^{-2}\alpha -1.6\times {10}^{-2}=0$

By applying quadratic equation formula,

$\alpha =\frac{[–b\pm \sqrt{(b{)}^2-4ac}]}{2a}$

On comparing with general quadratic equation $a{\alpha }^2+b\alpha +c=0$

$a=1,b=1.6\times {10}^{-2}\text{and}c=-1.6\times {10}^{-2}$

$\alpha =\frac{[–1.6\times {10}^{-2}\pm \sqrt{(1.6\times {10}^{-2}{)}^2-4(1\left)\right(-1.6\times {10}^{-2})}]}{2}$

$\alpha =\frac{[–1.6\times {10}^{-2}\pm \sqrt{2.56\times {10}^{-4}+6.4\times {10}^{-2}}]}{2}$

$\alpha =\frac{[–1.6\times {10}^{-2}\pm 2.53\times {10}^{-1}]}{2}$


The two values of the $\alpha$ are:

$\alpha =0.13$ and $–0.13$

The negative root is not acceptable and hence, $\alpha =0.13$

This means that the degree of dissociation $\left(\alpha \right)=0.13$.


Step 2: Equilibrium concentrations:

The degree of dissociation $\left(\alpha \right)=0.13$ {from step 1}

Form the equation (1),

$\left[H^{+}\right]\text{or}\left[H_3{O}^{+}\right]=\left[F^{-}\right]=c\alpha =0.02\times 0.13=2.6\times {10}^{-3}M$

$\left[HF\right]=c(1-\alpha )=0.02(1-0.13)=17.4\times {10}^{-3}M$


Step 3: 𝑝𝐻 of solution

$\left[H^{+}\right]=2.6\times {10}^{-3}M$ {from step 2}

We know $pH=-\log \left[H^{+}\right]$

$pH=-\log (2.6\times {10}^{-3})$

$pH=-\log \left(2.6\right)-\log \left({10}^{-3}\right)\left\{\log \right(a\times b)=\log a+\log b\}$

$pH=-0.42+3\left\{\log \right(a{)}^b=b\log a\}$

$pH=-0.42+3$

$pH=2.58$


Final answer:

(i) Degree of dissociation $=0.13$

(ii) Concentrations of species, $\left[H^{+}\right]=\left[F^{-}\right]=2.6\times {10}^{-3}M$ and $\left[HF\right]=17.4\times {10}^{-3}M$

(iii) pH of solution $=2.58$