Problem 3.12 : The solubility product of
AgBr is 5.2 x 10-13. Calculate its solubility
in mol dm' and g dm'(Molar mass of AgBr
= 187.8 g mol-')
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The solubility product of AgBr is 5.2 × 10¯¹³. molar mass of AgBr is 187.8 g/mol.
We have to find the solubility in mol/dm³ and g/dm³.
dissociation reaction of AgBr is ..
AgBr ⇔ Ag⁺ + Br¯
S S
so, solubility product, Ksp = [Ag⁺][Br¯]
⇒Ksp = S × S
⇒Ksp = S²
given, Ksp = 5.2 × 10¯¹³
∴ 5.2 × 10¯¹³ = S²
⇒S = √(5.2 × 10¯¹³) = 7.2 × 10^-7 mol/dm³
Therefore the solubility of AgBr is 7.2 × 10^-7 mol/dm³
∵ molar mass of AgBr is 187.8 g/mol
∴ the solubility of AgBr in g/dm³ = molar mass × solubility in mol/dm³
= 187.8 × 7.2 × 10^-7 g/dm³
= 1.352 × 10¯⁴ g/dm³
Therefore the solubility of AgBr in g/dm³ is 1.352 × 10¯⁴ g/dm³.
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