SATHEE: Equilibrium - Result Question 75

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) $4 : 5$

(d) $5 : 4$

[NEET Kar. 2013]

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Solution:

$∴ p K_{a} = - \log (1 \times 10^{- 4}) = 4$

Now from Handerson equation

$p H = p K_{a} + \log \frac{[Salt]}{[Acid]}$

Putting the values

$5 = 4 + \log \frac{[Salt]}{[Acid]}$

$\log \frac{[Salt]}{[Acid]} = 5 - 4 = 1$

Taking antilog

[Salt]/[Acid] $= 10 = 10 : 1$

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Equilibrium - Result Question 75

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