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7.59 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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Let the degree of ionization of propanoic acid be $\alpha$ . Then Let suppose we can write propanoic acid to be HA,

It is known that,

We have
ionization constant of propanoic acid ( $K_a$ )= $1.32\times 10^{-5}$ and the concentration is 0.005 M

$\alpha = \sqrt{\frac{K_a}{C}}$
By putting the values in above formula we get,

$\alpha = \sqrt{\frac{1.32\times 10^{-5}}{0.05}} = 1.62\times 10^{-2}$
[Acid] = $[H_{3}O]^+$ = C. $\alpha$ = $8.15\times 10^{-4}$

Therefore, $p^H = -\log [H_{3}O^+]$
$\\= -\log [8.15\times 10^{-4}]$
$=3.08$

If we add 0.01M hydrochloric acid then,

$AH+H_{2}O\rightleftharpoons H_{3}O^+ +A^-$
initial con C 0 0
at equi. C - $x$ $\approx$ C 0.01 + $x$ $x$

Now, by using the formula of $K_a = \frac{(x)(0.01)}{C-x}$
$=\frac{(x)(0.01)}{C}$

The value of $x$ is calculated as ;
$\Rightarrow 1.32\times 10^{-5}\times \frac{0.01}{0.01} = 1.32 \times 10^{-5}$ (Degree of ionisation)

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