1
Question
we have taken a saturated solution of AgBr. Ksp of AgBr is 12×10−4 if 10−7 mole of AgNO3 are added to 1 litre of this solution then the conductivity of solution in terms of 10−7Sm−1 units will be ? GIVEN λAg+=4×10−3;λ Br_= 6×10−3Sm2m
Open in App
Solution
AgBr=Ag++Br−
S + 10−7 S
AgNO3→Ag++NO−3
S + 10^-7 S
The solubility of AgBr in presence of 10−7molarAgNO3is,S=3×10−7 M.
Therefore, [Br-] = S = 43×10−7M, [Ag+] = S + 10−7=4×10−7Mand[NO−3]=S=10−7 M
Therefore, Ksolution = KBr−+KAg++KNO3−=[(6×10−3×3×10−7)+(4×10−3×4×10−7)+(5×10−3×10−7)]×1000=39Sm−1
AgBr=Ag++Br−
S + 10−7 S
AgNO3→Ag++NO−3
S + 10^-7 S
The solubility of AgBr in presence of 10−7molarAgNO3is,S=3×10−7 M.
Therefore, [Br-] = S = 43×10−7M, [Ag+] = S + 10−7=4×10−7Mand[NO−3]=S=10−7 M
Therefore, Ksolution = KBr−+KAg++KNO3−=[(6×10−3×3×10−7)+(4×10−3×4×10−7)+(5×10−3×10−7)]×1000=39Sm−1
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
