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Standard XII

Chemistry

Hydrogen Peroxide

0.1 M solutio...

Question

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

0.11, 0.09

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0.1, 0.1

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0.09, 0.11

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0.01, 0.01

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Solution

The correct option is C
0.09, 0.11

Initial pOH

pOH = $p K_{b}$ + log $\frac{[N H_{4} C l]}{[N H_{3}]}$ = $p K_{b}$ + log $\frac{0.1}{0.1}$ = $p K_{b}$

Number of equivalents of $N H_{3}$ = $N \times V_{l i t} = 0.1 \times 1$ = 0.1 eq
Number of equivalents of $N H_{4} C l$ = $N \times V_{l i t} = 0.1 \times 1$ = 0.1 eq
Number of equivalents of $H C l$ added = 0.01 eq

$N H_{3} + H C l \to N H_{4} C l$

Initially [ $N H_{3}$ ] = 0.1, [ $H C l$ ] = 0.01, [ $N H_{4} C l$ ] = 0.1
Finally [ $N H_{3}$ ] = 0.1 – 0.01 = 0.09, [ $H C l$ ] = 0, [ $N H_{4} C l$ ] = 0.1 + 0.01 = 0.11

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0.1 M solution of each of NH3 and NH4Cl 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 NH3 and NH4Cl are

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