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Potassium Permanganate

1 lit of a bu...

Question

1 lit of a buffer solution contains 0.1 mole each of $N H_{4} O H$ and $N H_{4} C l$ . What will be the $p^{H}$ of the solution when 0.01 mole of HCl is added to it?
$[P^{K b} o f N H_{4} O H = 4.74]$

A

9.26

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B

9.17

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C

4.74

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D

4.65

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Solution

The correct option is A 9.17
Initially, $[N H_{4} O H] = 0.1 m o l e$ and $[N H_{4} C l] = 0.1 m o l e$
When $H C l$ is added in the solution, then
$N H_{4} O H + H C l \to N H_{4} C l + H_{2} O$
Thus, 0.01 mole of $N H_{4} O H$ is used to dissociate $H C l$
thus, $[N H_{4} O H] = 0.1 - 0.01 = 0.99$
and $[N H_{4} C l] = 0.1 + 0.01 = 0.11$
Now, According to formula of buffer solution
pOH = pKb + log(0.11/0.99)
= 4.82
pH= 9.17

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