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Basic Buffer Action

The amount of...

Question

The amount of $(N H_{4})_{2} S O_{4}$ to be added to $500$ mL of $0.01$ m $N H_{4} O H$ solution ( $p K_{a}$ for $N H_{4}^{+}$ is $9.26$ ) prepare a buffer of $p H$ $= 8.26$ is?

A

0.05

mole

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B

0.025

mole

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C

0.10

mole

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D

0.005

mole

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Solution

The correct option is C $0.10$ mole
We are given, $p H = 8.26$
$p O H = 14 - p H$
$= 14 - 8.26$
$= 5.74$
Also, $p K_{a}$ $= 9.26$
$p K_{b} = 14 - p K_{a}$
$= 14 - 9.26$
$= 4.74$
Using Henderson Hasselbalch equation,
$p O H = p K_{b} + l o g \frac{[N H_{4}^{+}]}{[N H_{4} O H]}$
$5.74 = 4.74 + l o g \frac{[N H_{4}^{+}]}{0.01}$
$1 = l o g [N H_{4}^{+}] - l o g (0.01)$
$1 = l o g [N H_{4}^{+}] + 2$
$l o g [N H_{4}^{+}] = - 1$
Taking antilog
$[N H_{4}^{+}] = 0.1$

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Q. To prepare a buffer of pH = 8.26, amount of

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500

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