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Henderson Hassleback Equation

To 1L solutio...

Question

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 $(p K_{b} f o r N H_{3} = 4.74)$

-0.48

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0.48

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0.30

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-0.30

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Solution

The correct option is B
0.48

$N H_{4} C l + N a O H \to N a C l + N H_{3} + H_{2} O M o l e s o f N H_{4}^{+} l e f t = 0.1 - 0.05 = 0.05$
Total moles of $N H_{3} = 0.1 - 0.05 = 0.15$
$(p O H)_{1} = p H_{b} + l o g \frac{[S a l t]}{[B a s e]} = p K_{b} + l o g \frac{0.1}{0.1} = p K_{b}$
$(p O H)_{2} = p H_{b} + l o g \frac{0.05}{0.15} = p K_{b} - l o g 3$
Change in $p O H = (p O H)_{2} - (p O H)_{1} = - l o g 3$
Change in pH = log3 = 0.48

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0.10

N H_{3}

N H_{4} C l

(i)

0.02

H C l

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(K_{b} f o r N H_{3} = 1.8 \times 10^{- 5})

When 0.1 mole solid NaOH is added in 1L of 0.1 M $N H_{3}$ (aq.) $(K b = 2 \times 10^{- 5})$ , then select the correct statement(s):

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C H_{3} K C O O N a

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{N H}_{4} C l

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N H_{4} O H

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p^{H}

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