wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The sparingly soluble salt SrF2 has a Ksp value of 2.45×109 at room temperature. Calculate its solubility in 0.1 M highly soluble NaF aqueous solution.

A
2.45×107 molL1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
9×109 molL1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
4×1010 molL1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2.9×108 molL1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 2.45×107 molL1
Lets consider solid SrF2 is in contact with its saturated aqueous solution.The equilibrium is established between undissolved SrF2 and its ions. The equilibrium reaction is given by :
SrF2 Sr2++2Ft=teq cs s 2s

When NaF is added , as it is a strong electrolyte , So it completely ionized. It shall provide [F] ion concentration i.e. [F]=0.1 M

[Sr2+]=s[F]=2s+0.1

Solubility product Ksp=[Sr2+][F]2
Since NaF is a strong electrolyte , so it completely dissociates into ions. 2s+0.10.1

Ksp=[Sr2+][F]2
2.45×109=s(0.1)2
s=2.45×1090.01
s=2.45×107 mol L1

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