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JEE Main & Advanced Sample Paper JEE Main Sample Paper-43

  • question_answer
    To 100 mL of 0.1 M solution of sodium dihydrogen phosphate 75 mL of 0.1 M sodium phosphate is added. Calculate the pH when 25 mL of 0.1 M HCI is added to the above solution (stepwise acid dissociation constant for phosphoric acid are \[{{10}^{-3}},\,\,{{10}^{-6}}\] and \[{{10}^{-13}}\])

    A) 6

    B) 4

    C) 8.5

    D) None of these

    Correct Answer: D

    Solution :

    \[\Rightarrow \] \[v'\frac{v}{\sqrt{2}}\] 75 mL of 0.1 M \[E=\frac{1}{2}mv{{'}^{2}}+\frac{1}{2}mv{{'}^{2}}+\frac{1}{2}(2m){{v}^{2}}\] added       \[=mv{{'}^{2}}+m{{v}^{2}}\] 25 ml of 0.1 M HCl added \[\Rightarrow \] M moles          \[m{{\left( \frac{v}{\sqrt{2}} \right)}^{2}}+m{{v}^{2}}=\frac{3}{2}m{{v}^{2}}\]          \[{{T}_{1}}-{{T}_{2}}=12a\]     0          0 Added                   75                    2.5 After reaction 7.5 - 2.5               0          25     = 5 It forms buffer solution of \[{{T}_{2}}=3a\] and \[{{T}_{2}}\] M moles \[Eq\]      2.5 = 10 \[{{T}_{1}}=15a\] \[\therefore \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{15a}{3a}=\frac{5}{1}\] \[\Rightarrow \]


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