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Question
The solubility of AgBr in water is 1.20 × 10–5 mol dm–3. Calculate the solubility product of AgBr.
Solution
Given: Solubility of AgBr = 1.20 × 10−5
To find: Solubility product of AgBr
Formula: Ksp = xx yy Sx+y
Calculation: The solubility equilibrium of AgBr is:
\[\ce{AgBr_{(s)} <=> Ag^+_{ (aq)} + Br^-_{ (aq)}}\]
x = 1, y = 1
Ksp = xx yy Sx+y = (1)1 (1)1 S1+1 = S2
Ksp = (1.20 × 10-5)2
= 1.44 × 10-10
Solubility product of AgBr is 1.44 × 10-10
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