Chemical And Ionic Equilibrium

Chemical and Ionic Equilibrium - Result Question 137

79. How many moles of sodium propionate should be added to $1 L$ of an aqueous solution containing 0.020 mole of propionic acid to obtain a buffer solution of $pH $ $4.75$ ? What will be $pH$ if $0.010$ moles of $HCl$ are dissolved in the above buffer solution? Compare the last $pH$ value with the $pH$ of $0.010$ $ M $ $HCl$ solution. Dissociation constant of propionic acid, $K _a$ at $25^{\circ} C$ is $1.34 \times 10^{-5}$.

(1981, 4M)

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Solution:

For acidic buffer, the Henderson’s equation is

$pH=p K _a+\log \frac{(\text { mole of salt })}{(\text { mole of acid })} $

$4.75=-\log \left(1.34 \times 10^{-5}\right)+\log \frac{x}{0.02}$

$\Rightarrow x=0.015$ mole of sodium propionate. Addition of $0.01$ mole HCl will increase moles of propionic acid by $0.01$ and moles of sodium propionate will decrease by same amount.

New moles of acid $=0.02+0.01=0.03$

New moles of salt $=0.015-0.01=0.005$

$\mathrm{pH}=-\log \left(1.34 \times 10^{-5}\right)+\log \left(\frac{0.005}{0.030}\right)=4.09$

pH of $0.01 $ $\mathrm{HCl}=2$, just half of the pH of final buffer solution.