JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Sample Paper Topic Test - Ionic Equilibrium

  • question_answer
    Match List I with List II and select the correct answer using the code given below the lists
    List I List II
    [A] pH of 0.1 M HA \[(p{{K}_{a}}=5)\] and 0.01 M NaA (p) 4
    [B] pH of 0.1 M BOH \[(p{{K}_{b}}=6)\] and 0.1 M BCI (q) 7
    [C] pH of 0.1 M salt of HA \[(p{{K}_{a}}=5)\] and BOH \[(p{{K}_{b}}=7)\] (r) 6
    [D] pH of 500 litre of 0.2 M \[HN{{O}_{3}}\] and 500 litre 0.2 M NaOH (s) 8
    Codes:

    A) A\[\to \]p, B\[\to \]s, C\[\to \]r, D\[\to \]q               

    B) A\[\to \]s, B\[\to \]p, C\[\to \]r, D\[\to \]q

    C) A\[\to \]p, B\[\to \]r, C\[\to \]s, D\[\to \]q               

    D) A\[\to \]p, B\[\to \]s, C\[\to \]q, D\[\to \]r

    Correct Answer: A

    Solution :

    [a] \[pH=p{{K}_{a}}+{{\log }_{10}}\left[ \frac{\text{Salt}}{\text{Acid}} \right]=\]\[5+{{\log }_{10}}\left[ \frac{0.1}{0.01} \right]=4\]
    [b] \[pOH=p{{K}_{b}}+{{\log }_{10}}\left[ \frac{0.1}{0.1} \right]=6\]
    \[pH=14-6=8\]
    [c] \[pH=\frac{1}{2}[p{{K}_{w}}+p{{K}_{a}}-p{{K}_{b}}]=\]
    \[\frac{1}{2}[14+5-7]=6\]
    [d] Neutral solution \[pH=7\]


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