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Byju's Answer

Standard XII

Chemistry

Buffer Solutions

In a buffer s...

Question

In a buffer solution, the ratio of the concentration of $N H_{4} C I$ and $N H_{4} O H$ is 1 : 1 when it changes to 2 : 1, what will be the effect on pH of the buffer?

Increase

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Decrease

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No change

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First increase then decrease

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Solution

The correct option is B Decrease
in a basic buffer ,
$p O H = p K_{b} + l o g \frac{[s a l t]}{[b a s e]}$
$∴$ increasing Base i.e. $N H_{4} C l$ will increase $p O H$ and hence $p H$ will decrease.

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Similar questions

Q. In a buffer solution consisting of a weak acid and its salt, the ratio of concentration of salt to acid is increased tenfold, then the

p H

of the solution will:

Q. A buffer solution with pH 9 is to be prepared by mixing

N H_{4} C I

and

N H_{4} O H

. Calculate the number of moles of

N H_{4} C I

that should be added to 1 liter of 1M

N H_{4} O H

K_{b} (N H_{4} O H) = 1.8 * 10^{- 5}

Q. Concentration of

N H_{4} C l

and

N H_{4} O H

in a buffer solution are in the ratio of 1:10, Kb for

N H_{4} O H

10^{- 10}

. The pH of the buffer is:
(BHU, 1994)

Q. A buffer solution with

p H = 9

is to be prepared by mixing

N H_{4} C I

and

N H_{4} O H

. Calculate the number of moles of

N H_{4} C I

that should be added to one litre pf

1.0 M

N H_{4} O H

[K_{b} = 1.8 \times 10^{- 5}]

Q. Buffer solution

A

of a weak monoprotic acid and its sodium salt in the concentration ratio

x : y

has

p H = (p H)

,. Buffer solution

B

of the same acid and its sodium salt in the concentration ratio

y : x

has

p H = (p H)_{2}

. If

(p H)_{2} - (p H)_{1} = 1

unit and

(p H)_{1} + (p H)_{2} = 9.5 u n i t s

, then:

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In a buffer solution, the ratio of the concentration of NH4CI and NH4OH is 1 : 1 when it changes to 2 : 1, what will be the effect on pH of the buffer?

The correct option is B Decreasein a basic buffer , pOH=pKb+log[salt][base] ∴ increasing Base i.e. NH4Cl will increase pOH and hence pH will decrease.

In a buffer solution, the ratio of the concentration of $N H_{4} C I$ and $N H_{4} O H$ is 1 : 1 when it changes to 2 : 1, what will be the effect on pH of the buffer?

The correct option is B Decrease
in a basic buffer ,
$p O H = p K_{b} + l o g \frac{[s a l t]}{[b a s e]}$
$∴$ increasing Base i.e. $N H_{4} C l$ will increase $p O H$ and hence $p H$ will decrease.