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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    0.365 g of HCl gas was passed through 100 \[c{{m}^{3}}\]of\[0.2\text{ }M\text{ }NaOH\]solution. The pH of the resulting solution would be:

    A)  1                                            

    B)  5

    C)  8                            

    D)         9

    E)  13

    Correct Answer: E

    Solution :

    Moles of \[HCl=0.365\] Moles of \[NaOH=100\times 0.2=20\] Moles of\[NaOH\]left\[=19.635=20\] Moles of\[NaOH\]per litre \[=\frac{20\times {{10}^{-3}}\times 1000}{1000}\]                 \[pOH=-\log (2\times {{10}^{-2}})\]                 \[=1.7000=1.000\]                 \[pOH=1\]                 \[pH=14-1=13\]


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