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Standard XII

Chemistry

Basic Buffer Action

How many mole...

Question

How many moles of sodium hydroxide can be added to $1 L$ of a solution, $0.1 M$ in ${N H}_{3}$ and $0.1 M$ in ${N H}_{4} C l$ without changing the $p O H$ by more than $2.00$ unit? Assume no change in volume. $K_{b} {(N H}_{3}) = 1.8 \times 10^{- 5}$ .

0.0818

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0.818

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0.00818

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8.18

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Solution

The correct option is B $0.0818$
Initial $p O H = p K_{b} = 4.744$
Let $x$ mole of $N a O H$ has been added.
$[{N H}^{+}] = 0.1 + x, [{N H}_{3}] = 0.1 - x$
$p O H = 5.744$
$5.744 = 4.744 + log \frac{0.1 + x}{0.1 - x}$
$1 = log \frac{0.1 + x}{0.1 - x}$
$\frac{0.1 + x}{0.1 - x} = 10$ $\Rightarrow$ $x = \frac{0.9}{11} = 0.0818$ moles

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Similar questions

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):

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

Q. To a solution of

0.1 M

M g^{2 +}

and

0.8 M

N H_{4} C l

, an equal volume of

N H_{3}

is added which just gives precipitate. Calculate

[N H_{3}]

in solution.

K_{s p}

M g (O H)_{2} = 1.4 \times 10^{- 11}

and

K_{b}

N H_{4} O H = 1.8 \times 10^{- 5}

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

0.10 m o l e

each of

{N H}_{3}

and

{N H}_{4} C l

upon addition of (i)

0.02

mole of dissolved gaseous

H C l

, (ii)

0.02 m o l e

of dissolved

N a O H

. Assume no change in volume

K_{b}

for

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

)

Q. To a solution of

0.1 M

{M g}^{2 +}

and

0.8 M

{N H}_{4} C l

, an addition of equal volume of

{N H}_{3}

gives precipitate.

[{N H}_{3}]

in solution is:

[

K_{{s p}_{M g {(O H)}_{2}}} = 1.4 \times 10^{- 11}

and

K_{b_{{N H}_{4} O H}} = 1.8 \times 10^{- 5}

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How many moles of sodium hydroxide can be added to 1L of a solution, 0.1M in NH3 and 0.1M in NH4Cl without changing the pOH by more than 2.00 unit? Assume no change in volume. Kb(NH3)=1.8×10−5.

The correct option is B 0.0818Initial pOH=pKb=4.744Let x mole of NaOH has been added.[NH+]=0.1+x,[NH3]=0.1−xpOH=5.7445.744=4.744+log0.1+x0.1−x1=log0.1+x0.1−x0.1+x0.1−x=10 ⇒ x=0.911=0.0818 moles

How many moles of sodium hydroxide can be added to $1 L$ of a solution, $0.1 M$ in ${N H}_{3}$ and $0.1 M$ in ${N H}_{4} C l$ without changing the $p O H$ by more than $2.00$ unit? Assume no change in volume. $K_{b} {(N H}_{3}) = 1.8 \times 10^{- 5}$ .