Equilibrium Question 840
Question: Zirconium phosphate $ [Zr _3{{( PO_4 )} _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 then which of the following relationship between S and $ K _{sp} $ is correc'
Options:
A) $ S={K _{sp}/{{(6912)}^{1/7}}} $
B) $ S={{{K _{sp}/144}}^{1/7}} $
C) $ S={({{K _{sp}/6912}})^{1/7}} $
D) $ S={{{K _{sp}/6912}}^{7}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ [Zr_3{{(PO_4)}_4}]\rightarrow \underset{3S}{\mathop{3Z{r^{4+}}}},+\underset{4S}{\mathop{4PO_4^{3-}}}, $
$ K _{sp}={{(3S)}^{3}}{{(4S)}^{4}} $
$ =27S^{3}\times 256S^{4} $
$ =6912S^{7}. $
$ \therefore S={{( \frac{K _{sp}}{6912} )}^{1/7}} $
