One litre of a buffer solution is prepared by dissolving 0-6 mole of NH 3 and 0-4 mole of NH 4 Cl- What is the pH of the solution- for NH 3- K b -1-85- 10 -5-i What is the pH of the buffer after addition of 0-1 mole of HCl-ii What is the pH of the buffer after addition of 0-1 mole of NaOH

pOH=p K _ b +log [Salt] [Base] #160; pOH=5 log(1.85)+log 0.4 0.6 #160; pOH=4.57 pH=14 4.57=9.43 (1)After addition of 0.1 mole of ...

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Standard XII

Chemistry

Basic Buffer Action

One litre of ...

Question

One litre of a buffer solution is prepared by dissolving $0.6$ mole of ${N H}_{3}$ and $0.4$ mole of ${N H}_{4} C l$ . What is the pH of the solution? for ${N H}_{3}$ , $K_{b} = 1.85 \times 10^{- 5}$ .
(i) What is the pH of the buffer after addition of $0.1$ mole of $H C l$ .
(ii) What is the pH of the buffer after addition of $0.1$ mole of $N a O H$

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Solution

$p O H = p K_{b} + log \frac{[S a l t]}{[B a s e]}$
$p O H = 5 - l o g (1.85) + log \frac{0.4}{0.6}$
$p O H = 4.57$
$p H = 14 - 4.57 = 9.43$

(1)After addition of 0.1 mole of $H C l$ ,0.1 mole of $N H_{3}$ reacts with 0.1 mole of $H C l$ to form 0.1 mole of $N H_{4} C l$ .
So, $[N H_{4} C l] = 0.5$
$[N H_{3}] = 0.5$
$p O H = 5 - l o g (1.85) + log \frac{0.5}{0.5}$
$p O H = 4.748$
$p H = 14 - 4.74 = 9.26$

(2)After addition of 0.1 mole of $N a O H$ ,0.1 mole of $N H_{4} C l$ reacts with 0.1 mole of $N a O H$ to form 0.1 mole of $N H_{3}$ .
So, $[N H_{4} C l] = 0.3$
$[N H_{3}] = 0.7$
$p O H = 5 - l o g (1.85) + log \frac{0.3}{0.7}$
$p O H = 4.38$
$p H = 14 - 4.38 = 9.62$

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One litre of a buffer solution is prepared by dissolving 0.6 mole of NH3 and 0.4 mole of NH4Cl. What is the pH of the solution? for NH3, Kb=1.85×10−5.(i) What is the pH of the buffer after addition of 0.1 mole of HCl.(ii) What is the pH of the buffer after addition of 0.1 mole of NaOH