A saturated solution of silver bromide is made in 10 –7 (M) AgNO 3 solution
Given :
K sp of AgBr = 3 × 10 –13 
(Ag
+ ) = 6 × 10 –3 S m 2 mol –1 
(NO 3
– ) = 7 × 10 –3 S m 2 mol –1 
(Br
– ) = 8 × 10 –3 S m 2 mol –1 For water, κ = 7.5 × 10
–6 S m –1 Determine the conductivity of solution × 10
7 in S m –1 in nearest possible integers
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
158
Sol. Let solubility of AgBr (s) in AgNO 3 soluiton
= s (M)
AgBr (s) 
Ag + + Br –
(s + 10 –7 ) s ∴ s (s + 10
–7 ) = 3 × 10 –13 or s
2 + 10 –7 s – 30 × 10 –14 = 0
or s 2 + 6 × 10 –7 s – 5 × 10 –7 s – 30 × 10 –14 = 0
or s(s + 6 × 10 –7 ) –5 × 10 –7 (s + 6 × 10 –7 ) = 0
or (s – 5 × 10 –7 ) (s + 6 × 10 –7 ) = 0
∴ s = 5 × 10 –7 (M)
(AgNO 3 ) = 
(Ag + ) + 
(NO 3 – )
= 13 × 10 –3 S m 2 mol –1 
(AgBr) = 
(Ag
+ ) + 
(Br – )
= 14 × 10 –3 S m 2 mol –1 κ (AgNO 3 ) = [ 
(AgNO 3 )] C
= 13 × 10
–7 S m –1 κ (AgBr) = [ 
(AgBr)] C
= 70 × 10
–7 S m –1 κ solution = κ (AgNO 3 ) + κ (AgBr) + κ (H 2 O)]
= ( 13 × 10
–7 + 70 × 10 –7 + 75 × 10 –7 )
= 158 × 10 –7 S m –1
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems