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JEE Main & Advanced Sample Paper JEE Main Sample Paper-26

  • question_answer
    An acidic buffer solution has \[[HA]=1.0\]M and\[[NaA]=1.0M\]. To \[(10+x)\]mL of this buffer solution 9 mL of \[1.0M\,HCl\] is added so that pH changes by one unit. The value of \[x\] is

    A)  \[0.1\]                          

    B)  10

    C)  \[1.5\]                                      

    D)  \[1.0\]

    Correct Answer: D

    Solution :

    For given buffer solution pH,=pKa Now 9 mL of 1 M HCl is added to (10 + x) mL of this solution \[NaA\,\,\,\,\,+\,\,\,\,\,HCl\,\xrightarrow{\,}\,\,\,HA\,\,\,\,\,+\,\,\,NaCl\] \[1\times (10+x)\,\] \[(1\times 9)\] \[1(10+x)\,\,\,\,0\] \[(10+x)-9\,\,\,0\]                        \[(10+x)\,+9\] \[p{{H}_{1}}-p{{H}_{2}}=1\] \[\log \,\frac{(10+x)-9}{(10+x)\,+9}=\frac{1}{10}\] \[\frac{(10+x)\,-9}{(10+x)\,+9}=\frac{1}{10}\] \[x=1\,\,ml\].


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