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 K sp then which of the following relationship between S and Ksp is correct ?

Question

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 K sp then which of the following relationship between S and Ksp is correct ? (A) S = Ksp/ (6912)1/7  (B) S= Ksp/144 1/7 (C) S = Ksp/6912 1/7 (D) S = Ksp/6912 7

Tardigrade Admin · Accepted Answer

Zr3(PO4)4 → underset3s3Zr+3+ underset4s4 PO-34 Ksp =(3 s )3(4 s )4 s =(( ksp /6912))1 / 7

$S = {K_{s p} / (6912)^{1/7}}$

$S = {K_{s p} /144}^{1/7}$

$S = {K_{s p} /6912}^{1/7}$

$S = {K_{s p} /6912}^{7}$

$Z r_{3} (P O_{4})_{4} \to 3 s 3 Z r^{+ 3} + 4 s 4 P O_{4}^{- 3}$

$Ks p = (3 s)^{3} (4 s)^{4}$

$s = (\frac{k s p}{6912})^{1/7}$

S={Ksp​/(6912)1/7}

S={Ksp​/144}1/7

S={Ksp​/6912}1/7

S={Ksp​/6912}7

Zr3​(PO4​)4​→3s3Zr+3​+4s4PO4−3​​ Ksp=(3s)3(4s)4 s=(6912ksp​)1/7

$S = {K_{s p} / (6912)^{1/7}}$

$S = {K_{s p} /144}^{1/7}$

$S = {K_{s p} /6912}^{1/7}$

$S = {K_{s p} /6912}^{7}$

$Z r_{3} (P O_{4})_{4} \to 3 s 3 Z r^{+ 3} + 4 s 4 P O_{4}^{- 3}$

$Ks p = (3 s)^{3} (4 s)^{4}$

$s = (\frac{k s p}{6912})^{1/7}$