MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Basic Buffer Action

Concentration...

Question

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 $N H_{4} O H$ is $10^{- 10}$ . The pH of the buffer is:
(BHU, 1994)

A

4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

5

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C

9

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

11

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 5
The expression for the pOH of the basic buffer solution is as given below.
$p O H = p K_{b} + l o g \frac{[S a l t]}{[B a s e]}$ .
$p K_{b} = - l o g K_{b} = - l o g (10^{- 10}) = 10$ .
Substitute values in the above equation.
$p O H = 10 + l o g \frac{1}{10} = 9$ .
$p H = 14 - p O H = 14 - 9 = 5$ .

Suggest Corrections

thumbs-up

1

Similar questions

Q. Calculate the pH of the buffer solution containing

0.15 m o l e

{N H}_{4} O H

0.25 m o l e

{N H}_{4} C l

K_{b}

{N H}_{4} O H

1.8 \times 10^{- 5}

p K_{b}

N H_{4} O H

4.74

. If the ratio of the buffer mixture

[N H_{4} C l] : [N H_{4} O H]

1 : 10

p H

of that buffer is:

Q. In what volume ratio NH4Cl and NH4OH having 1M concentration should be mixed to get a buffer solution of pH=9.26 (pK_b of NH4OH = 4.74?

Q. Freshly precipitated

A l

M g

hydroxides are stirred vigorously in a buffer solution containing

0.25 M

{N H}_{4} C l

0.05 M

{N H}_{4} O H

[{A l}^{3 +}]

[{M g}^{2 +}]

in solution are:
Given:

K_{b}

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

K_{{s p}_{A l {(O H)}_{3}}} = 6 \times 10^{- 32}

K_{{s p}_{M g {(O H)}_{2}}} = 8.9 \times 10^{- 12}

Q. Freshly precipitated aluminium and magnesium hydroxides are stirred vigorously in a buffer solution containing

0.05 {mol L}^{- 1}

N H_{4} O H

0.25 {mol L}^{- 1}

N H_{4} C l

. Calculate the concentration of magnesium ions in solution.

K_{s p} A l (O H)_{3} = 6 \times 10^{- 32}

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

K_{s p} M g (O H)_{2} = 6 \times 10^{- 10}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Buffer Solutions

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Basic Buffer Action

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon