MyQuestionIcon

1

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

Organic Compounds

The acid diss...

Question

The acid dissociation constant of $H_{2} S$ and $H S^{-}$ are $10^{- 9}$ and $10^{- 13}$ respectively. The pH of 0.1 M aqueous solution of $H_{2} S$ will be:

A

2

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

B

3

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

C

4

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

D

5

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

Open in App

Solution

The correct option is D 5
$K_{a}$ for $H_{2} S = 10^{- 9}$
$K_{a}$ for $H S^{-} = 10^{- 13}$
$p H = \frac{1}{2} [p k_{a} - l o g c]$
$= \frac{1}{2} [- l o g (10^{- 9}) - l o g 0.1]$
$\frac{1}{2} [9 + 1] = 5$

Suggest Corrections

thumbs-up

0

Similar questions

Q. The dissociation constant for

H_{2} S

H S^{-}

are respectively

10^{- 7} a n d 10^{- 13}

0.1 M

aqueous solution of

H_{2} S

Q. Consider a diprotic acid

H_{2} A

. The dissociation constants for

H_{2} A

H A^{-}

10^{- 6} a n d 10^{- 13}

respectively. Then, the pH of

0.25 M

aqueous solution of

H_{2} A

Q. Calculate the appoximate pH of

0.1 M

H_{2} S

K_{1}

K_{2}

H_{2} S

1.0 \times 10^{- 7}

1.3 \times 10^{- 13}

respectively at

25^{o} C

Q. Calculate approximate

p H

0.1 M

aqueous solution of

H_{2} S, K_{1} a n d K_{2}

H_{2} S

1.00 \times 10^{- 7}

1.3 \times 10^{- 13}

respectively at

25^{o} C

Q. A solution contains

0.1 M

H_{2} S

0.3 M

H C l

. Calculate the concentration of

S^{2 -}

{H S}^{-}

ions is solution. Given

K_{a_{1}}

K_{a_{2}}

H_{2} S

10^{- 7}

1.3 \times 10^{- 13}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Introduction

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Organic Compounds

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon