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Standard XII

Chemistry

Acidic Buffer Action

Calculate the...

Question

Calculate the amount of $N H_{3}$ and $m o l L^{- 1}$ required to premare a buffer solution of pH 9.0 when total concentration of buffering reagents is 0.6 mol $L^{- 1} . p K_{b} f o r N H_{3} = 4.7, l o g 2 = 0.30$

0.4, 0.4

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0.2, 0.4

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0.4, 0.2

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0.2, 0.2

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Solution

The correct option is C 0.4, 0.2
$p O H = - l o g K_{b} + l o g \frac{[S a l t]}{[b a s e]}$ $5 = 4.7 + l o g \frac{a}{b} \Rightarrow \frac{a}{b} = 2$ $∴ a = 2 b$
Given $a + b = 0.6$
$2 b + b = 0.6$
$∴ 3 b = 0.6$
Or $b = 0.2 m o l e, o r 0.2 \times 17 = 3.4 g / L$
$∴ a = 0.4 m o l e$
Or $0.4 \times 53.5 = 21.4 g / L$
Thus [Salt] = 0.4M and [Base] = 4.8

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Similar questions

Q. Calculate the amount of

N H_{3}

and

N H_{4} C l

required to prepare a buffer solution of

p H 9.0

when total concentration of buffering reagents is

0.6 m o l L^{-}

p K_{b}

for

N H_{3} = 4.7, log 2 = 0.30

Q. Calculate the concentration of

N H_{3} a n d N H_{4}^{+} C l^{-}

present in the buffer solution of pH = 9, when total concentration of buffering reagents is 0.6

m o l L^{- 1} .

Take

p K_{b}

for

N H_{3} = 4.7, l o g 2 = 0.3

Q. Calculate the concentration of

N H_{3} a n d N H_{4}^{+} C l^{-}

present in the buffer solution of pH = 9, when total concentration of buffering reagents is 0.6

m o l L^{- 1} .

Take

p K_{b}

for

N H_{3} = 4.7, l o g 2 = 0.3

Calculate the amount of $N H_{3}$ and $N H_{4} C l$ required to prepare a buffer solution of pH = 9.0, when the total concentration of buffering reagents is 0.6 moles/L. Given $p K_{b}$ of $N H_{3}$ = 4.7

Q. A buffer solution is prepared in which the concentration of

N H_{3}

0.30 M

and the concentration of

N H_{4}^{+}

0.20 M

. If the equilibrium constant,

K_{b}

for

N H_{3}

equals

1.8 \times 10^{- 5}

, what is the

p H

of this solution?

(log 2 / 3 = - 0.176)

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Calculate the amount of NH3 and molL−1 required to premare a buffer solution of pH 9.0 when total concentration of buffering reagents is 0.6 mol L−1.pKb for NH3=4.7,log2=0.30

The correct option is C 0.4, 0.2pOH=−logKb+log[Salt][base] 5=4.7+logab⇒ab=2 ∴a=2b Given a+b=0.6 2b+b=0.6 ∴3b=0.6 Or b=0.2 mole, or 0.2×17=3.4g/L ∴a=0.4 mole Or 0.4×53.5=21.4g/L Thus [Salt] = 0.4M and [Base] = 4.8

Calculate the amount of $N H_{3}$ and $m o l L^{- 1}$ required to premare a buffer solution of pH 9.0 when total concentration of buffering reagents is 0.6 mol $L^{- 1} . p K_{b} f o r N H_{3} = 4.7, l o g 2 = 0.30$