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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2004

  • question_answer
    In a basic buffer \[0.0025\text{ }mole\]of \[N{{H}_{4}}Cl\] and \[0.15\text{ }mole\]of \[N{{H}_{4}}OH\]are present. The pH of the solution will be: \[{{(}_{P}}{{K}_{b}}=4.74)\]

    A)  \[11.04\]            

    B)         \[10.24\]

    C)  \[6.62\]              

    D)         \[5.48\]

    Correct Answer: A

    Solution :

    From Handersons equation \[pOH{{=}_{p}}{{K}_{b}}+\log \frac{[salt]}{[base]}\] \[=4.75+\log \frac{0.0025/V}{0.15/V}=4.74+\log \frac{1}{60}\] \[=4.74-1.78=2.96\] \[\therefore \]                  \[pH=14-2.96=11.04\]


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