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Chemistry Question on Equilibrium Constant

Calculate solubility of Zr3(PO4)4(s) {Zr_3(PO_4)_4(s)} in terms of KspK_{sp} ?

A

S=(Ksp6912)1/7S = \left(\frac{K_{sp}}{6912}\right)^{1/7}

B

S=(Ksp6912)1/6S = \left(\frac{K_{sp}}{6912}\right)^{1/6}

C

S=(Ksp1728)1/6S = \left(\frac{K_{sp}}{1728}\right)^{1/6}

D

S=(Ksp1728)1/7S = \left(\frac{K_{sp}}{1728}\right)^{1/7}

Answer

S=(Ksp6912)1/7S = \left(\frac{K_{sp}}{6912}\right)^{1/7}

Explanation

Solution

Zr3(PO4)4(s)<=>3Zr+4(aq)\+4PO4−3(aq) t=0a00 at  eqma−S3S4S\begin{matrix}&Zr_{3}\left(PO_{4}\right)_{4}\left(s\right)& {<=>}&3Zr^{+4}\left(aq\right)&\+ &4PO_{4}^{-3}\left(aq\right)\\\ t=0&a&&0&&0\\\ at \; eqm&a-S&&3S&&4S\end{matrix} Ksp=(Zr+4)3(PO4−3)4 {K_{sp} = (Zr^{+4})^3 (PO_4{^{-3}})^4} Ksp=(3S)3(4S)4K_{sp} = (3S)^3 (4S)^4 Ksp=6912  S7K_{sp} = 6912 \; S^7 S=(Ksp6912)1/7S = \left(\frac{K_{sp}}{6912}\right)^{1/7}