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Grade 12Engineering Entrance Exams

Solubility product of silver bromide is 5.0 x 10-13. The quantity of Gl taken as 120 g of mol-1) to be added to 1 litre of 0.05 M sol precipitation of AgBr is ?

Profile image of Vicki Sharma
12 Years agoGrade 12
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2 Answers

Profile image of Saurabh Koranglekar
6 Years ago
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Profile image of Arun
6 Years ago
General expression of solubility product -
 
M_{x}X_{y}\rightleftharpoons xM^{p+}(aq)+yX^{2-}(aq)
 
 
(x.p^{+}=y.\bar{q})
 
- wherein
 
Its solubility product is 
 
 
K_{sp}=[M^{p+}]^{x}[X^{q-}]^{y}
 
 
 
 AgBr \rightleftharpoons Ag ^++ Br^-
 
K_{sp} \; \; of \; \; AgBr = [ Ag^+][Br^-]
 
5 \times 10 ^{-13 }= 0.05 [Br ^- ]
 
[Br ^-] = \frac{5 \times 10^{13}}{0.05}= 1 \times 10^{-11}M
 
Moles of KBr = = 1 \times 1 0^{-11}
 
Weight of KBr = = 1.2 \times 1 0^{-9}g 

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