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  • Zirconium phosphate [Zr3(PO4)4]  dissociates into three zirconium cations of charge +4 and four phosphate anions of charge -3. If molar solubility of zirconium

Zirconium phosphate [Zr3(PO4)4]  dissociates into three zirconium cations of charge +4 and four phosphate anions of charge -3. If molar solubility of zirconium phosphate is denoted by S and its solubility product by Ksp then which of the following relationship between S and Ksp   is correct ?

Option: 1

S = {Ksp/(6912)1/7}


Option: 2

S = {Ksp / 144}1/7


Option: 3

S = (Ksp / 6912)1/7


Option: 4

S = {Ksp / 6912}7


Answers (1)

best_answer

The general expression of solubility product -

M_{x}X_{y}\rightleftharpoons xM^{p+}(aq)+yX^{2-}(aq)

Its solubility product is 
 

K_{sp}=[M^{p+}]^{x}\:[X^{2-}]^{y}

So,

 [Zr_{3(PO_{4})_{4}}]\rightleftharpoons 3Zr^{4^{+}}+4PO_{4}^{3-}

Solubility of [Zr_{3}(PO_{4})_{4}] is S

K_{sp}=[Zr^{4+}]^{3}[PO_{4}^{3-}]^{4}

= (3S)^{3}\times (4S)^{4}

= 3^{3}\times 4^{4}\times S^{3+4}

KSP = 6912 \times S^{7}

Solubility S= (K_{sp}/6912)^{1/7}

Posted by

Rishabh

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