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Henderson Hassleback Equation

To 1.0 L so...

Question

To $1.0 L$ solution containing $0.1$ mol each of $N H_{3}$ and $N H_{4} C l, 0.05$ mol NaOH is added. The change in pH will be : $(p K_{a}$ for $C H_{3} C O O H = 4.74$ )

0.30

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0.48

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- 0.48

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Solution

The correct option is D $- 0.48$
Initial concentration of $[N H_{3}] = \frac{0.1}{1} = 0.1 M$
Similarly, $[N H_{4} C l] = 0.1 M$
According to Henderson equation-
$P^{O H} = k b + log \frac{[s a l t]}{[b a s e]}$
$\Rightarrow P^{O H} = 4.74 + log \frac{0.1}{0.1} = 4.74$
$∴ P^{H} = 14 - P^{O H} \Rightarrow 14 - 4.74 = 9.26$
When $0.05$ mole of $N a O H$ added $⟶$
$N a O H + N H_{4} C l ⟶ N a C l + N H_{4} O H$
$t = 0$ $0.05$ $0.1$ $0$ $0.1$
$t = t_{f}$ $0$ $0.05$ $0.05$ $0.15$
Now, ${P^{O H}}_{2} = 14 - 4.26 = 9.74$
Change in $P^{H} \Rightarrow P^{H} - P H_{2} = - 0.48$

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Similar questions

To 1L solution containing 0.1 mol each of $N H_{3}$ and $N H_{4} C l, 0.05$ mol of NaOH is added. The change in pH will be $(p K_{b} f o r N H_{3} = 4.74)$

Q. 100ml of 0.1 M NaOH is added to 100 ml of a 0.2 M

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To 1.0L solution containing 0.1 mol each of NH3 and NH4Cl,0.05 mol NaOH is added. The change in pH will be : (pKa for CH3COOH=4.74)

To $1.0 L$ solution containing $0.1$ mol each of $N H_{3}$ and $N H_{4} C l, 0.05$ mol NaOH is added. The change in pH will be : $(p K_{a}$ for $C H_{3} C O O H = 4.74$ )