Question

Calculate solubility of AgBr2 i,e silver bromide if Ksp= 3.3×10^-13

Answer 1

Hey!
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→ Find the solubility of Silver Bromide when Ksp ( Solubility product ) = 3.3 × 10^-13 .


Ans. → The solubility is calculated as follows :


★ AgBr -------------> Ag+ + Br-

$= > ksp = s \: \times s \\ \\ = > ksp = s { }^{2} \\ \\ = > \sqrt{ksp} = s$


$= > s = \sqrt{3.3 \times 10 {}^{ - 13} } \\ \\ = > s = 5.74 \times 10 {}^{ - 7}$


•°• Solubility = 5.74 × 10^-7


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Answer 2

Answer. = √3.3 × 10^-13

S = 5.7 × 10^-7