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Buffer Solutions

In order to p...

Question

In order to prepare a buffer of $P^{H}$ 8.26, the amount of $(N H_{4})_{2} S O_{4}$ required to be mixed with 1L of 0.1 M $N H_{3} (p K_{b} = 4.74)$ is:

5 mol

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0.5 mol

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10 mol

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1 mol

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Solution

The correct option is C 1 mol
$p H = 8.26$
$p O H = 14 - 8.26 = 5.74$
$p K_{b} = 4.74$
$p O H = p K_{b} + l o g \frac{[S a l t]}{[B a s e]}$
$5.74 = 4.74 + l o g \frac{[N H_{4}^{+}]}{[0.1]}$
$1 = l o g [N H_{4}^{+}] - l o g 10^{- 1}$
$0 = l o g [N H_{4}^{+}]$
$[N H_{4}^{+}] = 1$
$⟹ 1$ $m o l$ of $(N H_{4})_{2} S O_{4}$ has to added.

Suggest Corrections

Similar questions

p H = 8.26

(N H_{4})_{2} S O_{4}

0.1 (M) N H_{3} (a q),

p K_{b}

4.74

(N H_{4})_{2} S O_{4}

500

0.01

N H_{4} O H

p K_{a}

N H_{4}^{+}

9.26

p H

= 8.26

({N H}_{4})_{2} {S O}_{4}

{N H}_{4} O H

[p K_{a} ({N H}_{4}^{+}) = 9.26]

C H_{3} C O O H

(0.1 N)

C H_{3} C O O N a

(0.1 N)

p H

p K_{a}

C H_{3} C O O H

p K_{a} (N H_{4}^{+}) = 9.2

(N H_{4})_{2} S O_{4}

N H_{4} O H

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