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JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Morning)

  • question_answer
    20 mL of 0.1 M \[{{H}_{2}}S{{O}_{4}}\] solution is added to 30 mL of 0.2 M \[N{{H}_{4}}OH\] solution. The pH of the resultant mixture is: \[[p{{K}_{b}}\text{ }of\text{ }N{{H}_{4}}OH\,\,=\,\,4.7]\]. [JEE Main Online Paper (Held On 09-Jan-2019 Morning]

    A) 5.2

    B) 9.4

    C) 9.0

    D) 5.0

    Correct Answer: C

    Solution :

    \[{{H}_{2}}S{{O}_{4}}+2N{{H}_{4}}OH~\,\,\to \,\,{{(N{{H}_{4}})}_{2}}S{{O}_{4}}\,+\,2{{H}_{2}}O\] \[t\,\,=\,\,0\]   2 millmole 6 millimole \[t\text{ }=\text{ }t\]    0     2 millimole   2 millimole \[\therefore \,\,\,\,\,\,\,\,{{P}^{OH}}\,=\,\,{{P}^{kb}}\,\,+\,\,\log \,\,\frac{[Salt]}{[Base]}\] \[{{P}^{OH}}\,=\,\,4.7\,\,+\,\,\log \,\,\frac{4/50}{2/50}\] \[\therefore \,\,{{P}^{OH}}\,=\,4.7\,\,+\,\,0.3\,\,=\,\,5\] \[\therefore \text{ }\,{{P}^{H}}\text{ }=\text{ }9.\]


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