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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Self Evaluation Test - Equilibrium

  • question_answer
    Zirconium phosphate \[[Z{{r}_{3}}{{\left( P{{O}_{4}} \right)}_{4}}]\] 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 correct?                    

    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] \[[Z{{r}_{3}}{{(P{{O}_{4}})}_{4}}]\rightleftharpoons \underset{3S}{\mathop{3Z{{r}^{4+}}}}\,+\underset{4S}{\mathop{4P{{O}_{4}}^{3-}}}\,\] \[{{K}_{sp}}={{(3S)}^{3}}{{(4S)}^{4}}\] \[=27{{S}^{3}}\times 256{{S}^{4}}\] \[=6912{{S}^{7}}.\] \[\therefore S={{\left( \frac{{{K}_{sp}}}{6912} \right)}^{1/7}}\]


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