If K sp for HgSO 4 is 6.4 × 1 0 − 5 , then solubility of this substance in mole per m 3 is

H g S O 4 ( s ) → H g 2 + ( a q ) + S O 4 2 − ( a q ) solubility, s = 6.4 × 1 0 − 5 = 8 × 1 0 − 3 mol / L = 8 mol / m 3 ]

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ChemistryIonic EquilibriumSolubility ProductEasy2 minQB

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If $K_{sp}$ for $HgSO_{4}$ is $6.4 \times 1 0^{- 5}$ , then solubility of this substance in mole per $m^{3}$ is

Answer:

Solution:

$H g S O_{4} (s) \to H g^{2 +} (a q) + S O_{4}^{2 -} (a q)$

solubility, $s = 6.4 \times 1 0^{- 5} = 8 \times 1 0^{- 3} mol / L = 8 mol / m^{3}]$

Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

⏱ 2mℹ️ Source: QB

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