The dissociation constant of a weak acid is $1 \times 10^{-4}$. In order to prepare a buffer solution with a…

The dissociation constant of a weak acid is $1 \times 10^{-4}$. In order to prepare a buffer solution with a $\mathrm{pH}=5$ the [Salt] $/[$ Acid $]$ ratio should be
  1. $1: 10$
  2. $4: 5$
  3. $10: 1$
  4. $5: 4$

Solution

$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { Salt }]}{[\text { Acid }]}$
$5=4+\log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]} \quad\left[\because \mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}ight]$
Given, $\mathrm{K}_{\mathrm{a}}=1 \times 10^{-4}$
$\therefore \mathrm{pK}_{\mathrm{a}}=-\log \left(1 \times 10^{-4}ight)=4$
Now from Handerson equation $\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { Salt }]}{[\mathrm{Acid}]}$
Putting the values $5=4+\log \frac{[\text { Salt }]}{[\text { Acid }]}$
$\log \frac{[\mathrm{Salt}]}{[\text { Acid }]}=5-4=1$
Taking antilog $[$ Salt $] /[$ Acid $]=10=10: 1$ ~

Asked in: JEE-TOPICTESTS-CHEMISTRY

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