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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 14

  • question_answer
    The concentration of \[C{{O}_{2}}\] in atmosphere is 88 ppm. If all of the \[C{{O}_{2}}\] present in \[{{10}^{5}}\text{ }mL\]of air is dissolved in \[1\text{ }d{{m}^{3}}\]water, then approximate \[pOH\]of solution at \[27{}^\circ C\]will be [\[{{K}_{a1}}={{10}^{-7}},\,{{K}_{a2}}={{10}^{-11}}\] for \[{{H}_{2}}C{{O}_{3}}\]]

    A) \[3.2\]      

    B)              \[3.85\]                  

    C) \[10.15\]                

    D)   None of these

    Correct Answer: C

    Solution :

    [c] 88 ppm means: \[{{10}^{6}}\text{ }mL\]of air contains \[=88\text{ }g\]of \[C{{O}_{2}}\] \[=\frac{88}{44}=2\,molC{{O}_{2}}\] Moles of \[C{{O}_{2}}\] is \[{{10}^{5}}mL=\frac{2}{{{10}^{6}}}\times {{10}^{5}}=0.2mol\] \[[C{{O}_{2}}]=\frac{0.2mol}{1d{{m}^{3}}}=0.2M\] Use direct reaction \[pH{{w}_{A}}=\frac{1}{2}(p{{K}_{a}}-\log C)\] \[=\frac{1}{2}(7-\log \,\,0.2)\] \[=\frac{1}{2}(7-0.3+1)\] \[=\frac{7.7}{2}\] \[=3.85\] \[=pOH=14-3.85=10.15\]


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