A) \[S=\{{{K}_{sp}}/{{(6912)}^{1/7}}\}\]
B) \[S={{\{{{K}_{sp}}/144\}}^{1/7}}\]
C) \[S={{\{{{K}_{sp}}/6912\}}^{1/7}}\]
D) \[S={{\{{{K}_{sp}}/6912\}}^{7}}\]
Correct Answer: C
Solution :
[c] \[[Z{{r}_{3}}{{(P{{O}_{4}})}_{4}}]\rightleftharpoons \underset{3S}{\mathop{3Z{{r}^{4+}}}}\,+\underset{4S}{\mathop{4P{{O}_{4}}^{3-}}}\,\] \[{{K}_{sp}}={{(3S)}^{3}}{{(4S)}^{4}}\] \[=27{{S}^{3}}\times 256{{S}^{4}}\] \[=6912{{S}^{7}}.\] \[\therefore S={{\left( \frac{{{K}_{sp}}}{6912} \right)}^{1/7}}\]You need to login to perform this action.
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