The solubility of Ba 3 ( AsO 4 ) 2 (molar mass = 690 ) is 6.9 × 1 0 − 2 g /100 mL . What is the value of its K sp ?

s = 6.9 × 1 0 − 2 g m /100 m l K s p = 690 6.9 × 1 0 − 2 = 1 0 − 4 m o l /100 m l = 1 0 − 3 m o l / l i t = 108 s 5 = 108 × ( 1 0 − 3 ) 5 = 108 × 1 0 − 15 = 1.08 × 1 0 − 13

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ChemistryIonic EquilibriumSolubility ProductMedium2 minQB

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The solubility of $Ba_{3} (AsO_{4})_{2}$ (molar mass $= 690$ ) is $6.9 \times 1 0^{- 2} g /100 mL$ . What is the value of its $K_{sp}$ ?

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Answer:

Solution:

$s = 6.9 \times 1 0^{- 2} g m /100 m l$

$K_{s p} = \frac{6.9 \times 1 0 ^{- 2}}{690} = 1 0^{- 4} m o l /100 m l = 1 0^{- 3} m o l / l i t = 108 s^{5} = 108 \times (1 0^{- 3})^{5} = 108 \times 1 0^{- 15} = 1.08 \times 1 0^{- 13}$

Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

⏱ 2mℹ️ Source: QB

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