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Henderson Hassleback Equation

To 1 L soluti...

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

A

-0.48

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B

0.48

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C

0.30

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D

-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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Similar questions

1.0 L

solution containing

0.1

N H_{3}

N H_{4} C l, 0.05

mol NaOH is added. The change in pH will be :

(p K_{a}

C H_{3} C O O H = 4.74

)

Q. Calculate the change in pH of one litre buffer solution containing

0.10 m o l e

{N H}_{3}

{N H}_{4} C l

upon addition of (i)

0.02

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H C l

0.02 m o l e

N a O H

. Assume no change in volume

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{N H}_{3} = 1.8 \times 10^{- 5}

Q. Calculate the amount of

N H_{3}

N H_{4} C l

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0.1 M solution of each of $N H_{3}$ and $N H_{4} C l$ are added to prepare a buffer of 1 litre solution. If 0.01 mole of HCl is added (assuming no volume change). The concentrations of $N H_{3}$ and $N H_{4} C l$ are

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