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JEE Main & Advanced Sample Paper JEE Main Sample Paper-46

  • question_answer
    \[1\,M\,N{{H}_{4}}\,OH\]and 1 M HCI are mixed to make total volume of 300 mL. If pH of the mixture is 9.26 and \[p{{K}_{a}}\,(NH_{4}^{+})=9.26\] then volume ratio of \[N{{H}_{4}}OH\] and HCI will be

    A)  1 : 1                           

    B)  1 : 2              

    C)  2 : 1                           

    D)  3 : 1

    Correct Answer: C

    Solution :

     pH = 9.26 shows that \[[N{{H}_{4}}OH]\,\,>\,HCl\] \[\therefore \] The mixture is a buffer. Let \[HCl=x\,\,mL=x\,\,mmol\] \[\therefore \] \[N{{H}_{4}}OH=\,(300-x)=\,(300-x)\,\,mmol\]and \[N{{H}_{4}}C{{l}_{formed}}=x\,\,mmol\] \[N{{H}_{4}}OH\] unreacted \[=300-x-x=(300-2x)\,m\,\,mol\] \[pOH=14-9.26=4.74\] \[p{{K}_{b}}=14-9.26=4.74\] \[pOH=p{{K}_{b}}+\log \,\frac{[NH_{4}^{+}]}{[N{{H}_{4}}OH]}\] \[4.74=4.74+\log \,\frac{x}{(300-2x)}\]\[\frac{x}{300-2x}=1\] \[\therefore \] \[x=100\,mL=\] Volume of HCl \[300-x=200\,mL=\] Volume of \[N{{H}_{4}}OH\] Hence, \[\frac{Volume\,of\,N{{H}_{4}}OH}{Volume\,of\,HCl}=\frac{200}{100}\]\[=\frac{2}{1}\] or 2 : 1


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