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Equilibrium — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium5 Marks
Question
The ionization constant of propanoic acid is $1.32 \times 10^{–5}$​​​​​​​. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?
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Answer

Let the degree of ionization of propanoic acid be α.
Then, representing propionic acid as HA, we have:
$\text{HA}$ $+$ $\text{H}_2\text{O}$ $\leftrightarrow$ $\text{H}_3\text{O}^+$ $+$ $\text{A}^-$
$(.05-0.0\alpha)\approx.05$ $.05\alpha$   $.05\alpha$
$\text{K}_\text{a}=\frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA]}}$
$=\frac{(.05\alpha)(.05\alpha)}{0.05}=.05\alpha^2$
$\alpha=\sqrt{\frac{\text{K}_\text{a}}{.05}}=1.63\times10^{-2}$
Then, $[\text{H}_3\text{O}^+]=.05\alpha=0.5\times1.63\times10^{-2}=\text{K}_\text{b}.15\times10^{-4}\text{M}$
$\therefore\ \text{pH}=3.09$
In the presence of 0.1M of HCl, let α´ be the degree of ionization.
$\text{Then, }[\text{H}_3\text{O}^+]=0.01$
$[\text{A}^-]=005\alpha'$
$[\text{HA]}=.05$
$\text{K}_\text{a}=\frac{0.01\times0.5\alpha'}{.05}$
$1.32\times10^{-5}=.01\times\alpha'$
$\alpha'=1.32\times10^{-3}$

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