Equilibrium - Result Question 75
####79. The dissociation constant of a weak acid is $1 \times 10^{-4}$. In order to prepare a buffer solution with a pH $=5$ the [Salt]/[Acid] ratio should be
(a) $1: 10$
(b) $4: 5$
(c) $10: 1$
(d) $5: 4$
[NEET Kar. 2013]
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Solution:
- (c) Given, $K_a=1 \times 10^{-4}$
$\therefore \quad p K_a=-\log (1 \times 10^{-4})=4$
Now from Handerson equation
$pH=p K_a+\log \frac{[\text{ Salt }]}{[\text{ Acid }]}$
Putting the values
$5=4+\log \frac{[\text{ Salt }]}{[\text{ Acid }]}$
$\log \frac{[\text{ Salt }]}{[\text{ Acid }]}=5-4=1$
Taking antilog
[Salt]/[Acid] $=10=10: 1$
