Question

I. How many moles of sodium propionate should be added to one litre of an aqueous solution contain $0.02$ mole of propionic acid to obtain a buffer solution of $pH$ $4.7$?
II. What will be the $pH$ if $0.0005$ mole of hydrogen chloride is dissolved in the above buffer solution?
III. Compare the last $pH$ value with the $pH$ of a $0.01$ molar $HCl$ solution.
Assume that dissociation constant of propionic acid, ${25}^{o}C$ is $1.00\times {10}^{-5}$

Answer 1

$I$ According to Henderson's Equation

$pH=p{K}_a+log\frac{Salt}{Acid}$

now $pH=4.70,$ $p{K}_a=-log{K}_a=-log\left({10}^{-5}\right)=5$

$4.70=5+log\frac{salt}{0.02}$

$Salt=100.24\times {10}^{-4}$ $mol/L$

number of moles of Sodium Proponate$=cone\times volume=100.24\times {10}^{-4}$ $mol/L$

$II$ Now $0.0005$ mole of $HCl$ is added to above buffer before adding $\left[H^{+}\right]$

$C_2{H}_{5}COOH\leftrightarrows H^{+}+C_2{H}_{5}CO{O}^{-}$

$K_a=\frac{\left[C_2{H}_{5}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C_2{H}_{5}COOH\right]}$

$\left[H^{+}\right]=\frac{{10}^{-5}\times 0.02}{100.24\times {10}^{-4}}=19.95\times {10}^{-6}\frac{mol}{L}$

	$C_2{H}_{5}CO{O}^{-}$	$H^{+}$	$C_2{H}_{5}COOH$
$Initial$ $Change$ $Equilibrium$	$100.24\times {10}^{-4}$ $-0.0005$ $0.009524$	$19.95\times {10}^{-6}$ $-0.0005$ $0$	$0.02$ $0.0005$ $0.0205$

$\left[H^{+}\right]=K_a=\frac{\left[H^{+}\right]\left[C_2{H}_{5}CO{O}^{-}\right]}{\left[C_2{H}_{5}COOH\right]}$

${10}^{-5}=\frac{\left[H^{+}\right][95.24\times {10}^{-4}]}{[205\times {10}^{-4}]}$

$\left[H^{+}\right]=21.5\times {10}^{-6}pH=4.6671$

$III$ The $pH$ would be $9.67$ after addition of $HCl$ on comparison of two values of $pH$, we get to know that the $pH$ of the buffer decreased on addition of $HCl$. the difference between two $pH$ value is $0.033$