Question

How much water must be added to $300\mathrm{ml}\mathrm{of}0.2\mathrm{M}$ solution of ${\mathrm{CH}}_3\mathrm{COOH}({\mathrm{K}}_{\mathrm{a}}=1.8\times {10}^{-5})$for the degree of dissociation of the acid to double?

Answer 1

Step 1: Given data:

${\mathrm{CH}}_3\mathrm{COOH}$as a weak acid dissociates in the aqueous solution as;

${\mathrm{CH}}_3\mathrm{COOH}\left(\mathrm{aq}\right)\leftrightharpoons {\mathrm{CH}}_3{\mathrm{COO}}^{–}\left(\mathrm{aq}\right)+{\mathrm{H}}^{+}\left(\mathrm{aq}\right)$

Where the dissociation constant is given as ;

${\mathrm{K}}_{\mathrm{a}}=\frac{[\mathrm{CH}3\mathrm{COO}-][\mathrm{H}+]}{\left[\mathrm{CH}3\mathrm{COOH}\right]}=1.8\times {10}^{-5}$

The concentration is given as ${\mathrm{C}}_1=0.2\mathrm{M}$

Also, the volume is given as,${\mathrm{V}}_1=300\mathrm{ml}$

Step 2: Calculation of Degree of dissociation using Ostwald dilution law:

We know from Ostwald dilution law that the dissociation constant of a weak acid is given as;
${\mathrm{K}}_{\mathrm{a}}={\mathrm{C\alpha }}^2$ (Where C is the concentration of acid, ${\mathrm{K}}_a$ is the dissociation constant and $\mathrm{\alpha }$ is the degree of dissociation)

Also;

$$\begin{gathered} \Rightarrow \mathrm{\alpha }=\sqrt{\frac{{\mathrm{K}}_{\mathrm{a}}}{\mathrm{c}}} \\ \Rightarrow \mathrm{\alpha }=\sqrt{\frac{1.8\times {10}^{-5}}{0.2}} \\ =9.486\times {10}^{-3} \end{gathered}$$

Now we double this value to get $\mathrm{\alpha };$

$$\begin{gathered} \mathrm{\alpha }=2\times 9.486\times {10}^{-3} \\ =18.97\times {10}^{-3} \end{gathered}$$

Step 3: Calculation of the new concentration:

To get the new concentration we get:

$$\begin{gathered} \Rightarrow 18.97\times {10}^{-3}=\sqrt{\frac{1.8\times {10}^{-5}}{{\mathrm{C}}_1}} \\ \Rightarrow 360\times {10}^{-6}=\frac{1.8\times {10}^{-5}}{{\mathrm{C}}_1} \\ \Rightarrow {\mathrm{C}}_1=\frac{1.8\times {10}^{-5}}{360\times {10}^{-6}}=0.05\mathrm{mol}/\mathrm{L} \end{gathered}$$

This is the new concentration we need.

Step 4: Calculation of volume of water needed:

To find how much water we need to add to achieve this dilution we can say:

$$\begin{gathered} {\mathrm{C}}_1{\mathrm{V}}_1={\mathrm{C}}_2{\mathrm{V}}_2 \\ \Rightarrow 300\times 0.2=0.05\times {\mathrm{V}}_2 \\ \Rightarrow {\mathrm{V}}_2=\frac{300\times 0.2}{0.05} \\ =1200\mathrm{ml} \end{gathered}$$

Hence, we need to add further $1200-300=900\mathrm{ml}$water to the solution.

Hence, water must be added to $300\mathrm{ml}\mathrm{of}0.2\mathrm{m}$ solutions of ${\mathrm{CH}}_3\mathrm{COOH}({\mathrm{K}}_{\mathrm{a}}=1.8\times {10}^{-5})$for the degree of dissociation of the acid to double is $900\mathrm{ml}$.