CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

The solubility product of a saturated solution of Ag2CrO4 in water at 298 K will be................., if the EMF of the cell:

Ag|Ag(satAg2CrO4sol)||Ag(0.1M)|Ag is 0.164V at 298 K.

A
2.287×1012M3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2.659×1012M3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1.965×1012M3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Noneofthese
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 2.287×1012M3

Cell circuit is given as below:

Ag|Ag (saturated) Ag2CrO4||Ag(0.1M)|Ag

Ecell=0.164 at 298 K.

For the given cell (concentration cell)

Ecell=0.0591nlogc2c1

c1=[Ag in Ag2CrO4 (left side)

c2=(Ag) (right side)

N = 1 (number of exchange electron)

So, 0.164=0.05911log0.1c1

Or log0.1c1=0.1640.0591 = 2.774

On solving

c1=[Ag]inAg2CrO4=1.66×104M

We get Ag2CrO42Ag+CrO24

[CrO24]=[Ag]2

So, [CrO24=1.66×1042M=0.83×104M

KspofAg2CrO4=[Ag]2[CrO24]

=(1.66×104)(0.83×104)

=2.287×1012M3

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Nernst Equation
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon