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Nernst Equation

To 1 litre ...

Question

To $1$ litre solution containing $0.1 m o l e$ each of $N H_{3}$ and $N H_{4} C l, 0.05 m o l$ of $N a O H$ is $p H$ will be $(p K_{b}$ for $N H_{3} = 4.74)$ .

1.48

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0.48

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0.20

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0.80

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Solution

The correct option is B $0.48$
Initial pH of the solution can be calculated as
$[N H_{3}] = \frac{N o . o f m o l e s}{V o l u m e} = \frac{0.1}{1} = 0.1 M$
Similarly, $[N H_{4} C l] = 0.1 M$
According to Henderson's equation
$p O H = K_{b} + log \frac{[s a l t]}{[b a s e]}$
$p O H = 4.74 + log \frac{0.1}{0.1}$
$p O H = 4.74$
Now, $p H = 14 - p O H = 14 - 4.74 = 9.26$
When o0.05 mole of NAOH is added
$N a O H + N H_{4} C l ⟶ N a C l + N H_{4} O H$
Initial moles 0.05 0.1 0 0.1
Final moles 0 0.05 0.05 0.15
Now, $p O H_{2} = 4.74 + log \frac{0.05}{0.15}$
$p O H_{2} = 4.74 + log 0.3333$
$p O H_{2} = 4.74 + 0.4771$
$= 4.26$
$p H_{2} = 14 - 4.74$
Change in $p H = p H - p H_{2} = - 0.48$
Thus we can say that the pH of the solution increases by 0.48

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