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Question
The solubility product of AgBr is 5.2 × 10−13 Calculate its solubility in mol dm−3 and g dm−3 (Molar mass of AgBr = 187.8g mol−1)
Sum
Solution
The solubility product of AgBr is:
AgBr(s) ⇌ Ag+(aq) + Br−(aq)
x = 1, y = 1
Ksp = [Ag+][Br−] = S2
`"S" = sqrt ("K"_(sp)) = sqrt (5.2 × 10^(−13))`
= 7.2 × 10−7 Mol dm−3
The solubility in g dm−3 = molar solubility in mol dm−3 × molar mass g mol−1
S = 7.2 × 10−7 mol dm−3 × 187.8 g mol−1
= 1.35 × 10−4 g dm−3
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Solubility Product - Solubility product
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