• 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

$\Rightarrow$ $v'\frac{v}{\sqrt{2}}$ 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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