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Manipal Engineering Manipal Engineering Solved Paper-2009

  • question_answer
    \[30\,\,cc\]of\[\frac{M}{3}HCl\],\[20\,\,cc\]of\[\frac{M}{2}HN{{O}_{3}}\]and\[40\,\,cc\] of\[\frac{M}{4}NaOH\]solutions are mixed and the volume was made up to \[\text{1}\,\,\text{d}{{\text{m}}^{3}}\]. The pH of the resulting solution is

    A)  8                                            

    B)  2

    C)  1                            

    D)         3

    Correct Answer: B

    Solution :

    Total milliequivalents of\[{{H}^{+}}\]                 \[=30\times \frac{1}{3}+20\times \frac{1}{2}=20\] Total milliequivalents of\[O{{H}^{-}}\]                 \[=40\times \frac{1}{4}=10\] Milli equivalence of \[{{H}^{+}}\] left                 \[=20-10=10\] \[\therefore \]  \[[{{H}^{+}}]=\frac{10}{1000}g\,\,ions/d{{m}^{3}}={{10}^{-2}}\] \[\therefore \]  \[pH=2\]


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