Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumSolubility ProductEasy2 minPYQ_2020

ChemistryEasysingle choice

Solubility of Zirconium phosphate $Z r_{3} (P O_{4})_{4}$ is 's' moles per litre. Solubility product of $K_{s p}$ may be given as

Answer:

Solution:

${Zr}_{3} {({PO}_{4})}_{4} ⇌ \underset{3 s}{3 {Zr}^{4 +}} + \underset{4 s}{4 {PO}_{4}^{3 -}}$
$K_{s p} = (3 s)^{3} \times (4 s)^{4} = 6912 s^{7}$

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

⏱ 2mℹ️ Source: PYQ_2020

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