The molar solubility (s) of zirconium phosphate with molecular formula is given by relations: [2025]

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Q.
The molar solubility (s) of zirconium phosphate with molecular formula ${({Zr}^{4 +})}_{3} {({PO}_{4}^{3 -})}_{4}$ is given by relations: [2025]

1 ${(\frac{K_{s p}}{9612})}^{\frac{1}{3}}$

2 ${(\frac{K_{s p}}{5348})}^{\frac{1}{6}}$

3 ${(\frac{K_{s p}}{8435})}^{\frac{1}{7}}$

4 ${(\frac{K_{s p}}{6912})}^{\frac{1}{7}}$

Ans.
(4)

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

$K_{s p} = {[Z r^{4 +}]}^{3} {[P O_{4}^{3 -}]}^{4} = {[3 s]}^{3} {[4 s]}^{4} = 6912 s^{7}$

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

Solution

Q.
The molar solubility (s) of zirconium phosphate with molecular formula ${({Zr}^{4 +})}_{3} {({PO}_{4}^{3 -})}_{4}$ is given by relations: [2025]

1 ${(\frac{K_{s p}}{9612})}^{\frac{1}{3}}$

2 ${(\frac{K_{s p}}{5348})}^{\frac{1}{6}}$

3 ${(\frac{K_{s p}}{8435})}^{\frac{1}{7}}$

4 ${(\frac{K_{s p}}{6912})}^{\frac{1}{7}}$

Ans.
(4)

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

$K_{s p} = {[Z r^{4 +}]}^{3} {[P O_{4}^{3 -}]}^{4} = {[3 s]}^{3} {[4 s]}^{4} = 6912 s^{7}$

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

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Solution

Q. The molar solubility (s) of zirconium phosphate with molecular formula (Zr4+)3 (PO43-)4 is given by relations: [2025]

1 (Ksp9612)13

2 (Ksp5348)16

3 (Ksp8435)17

4 (Ksp6912)17

Ans. (4) Zr3(PO4)4⇌3Zr4+3s+4PO43-4s Ksp=[Zr4+]3[PO43-]4=[3s]3[4s]4=6912s7 s=(Ksp6912)17

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Q.
The molar solubility (s) of zirconium phosphate with molecular formula ${({Zr}^{4 +})}_{3} {({PO}_{4}^{3 -})}_{4}$ is given by relations: [2025]

1 ${(\frac{K_{s p}}{9612})}^{\frac{1}{3}}$

2 ${(\frac{K_{s p}}{5348})}^{\frac{1}{6}}$

3 ${(\frac{K_{s p}}{8435})}^{\frac{1}{7}}$

4 ${(\frac{K_{s p}}{6912})}^{\frac{1}{7}}$

Ans.
(4)

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

$K_{s p} = {[Z r^{4 +}]}^{3} {[P O_{4}^{3 -}]}^{4} = {[3 s]}^{3} {[4 s]}^{4} = 6912 s^{7}$

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