A) 2.699
B) 13.30
C) 11.3010
D) 1.330
Correct Answer: C
Solution :
Given, volume of HC1 solution \[=200\text{ }mL\] \[pH\]of \[HCl\]solution \[=2.0\] \[\therefore \]\[[{{H}^{+}}]\]in \[\text{HCl}\]solution \[\text{=1}\times \text{1}{{\text{0}}^{-2}}\] \[\therefore \]Milliequivalents of \[\text{HCl}=200\times 1\times {{10}^{-2}}\] \[=2\] Similarly, volume of \[\text{NaOH}\]solution \[\text{=}\,\text{300}\,\text{mL}\] \[pH\]of \[\text{NaOH}\,\text{= 12}\] \[pOH\]of \[\text{NaOH}=14-12=2\] \[[O{{H}^{-}}]\]in \[\text{NaOH}=14-12=2\] \[[O{{H}^{-}}]\]in \[\text{NaOH}\]solution \[\text{=1}\times {{10}^{-2}}\] \[\therefore \]Milliequivalents of \[\text{NaOH}\] \[=300\times 1\times {{10}^{-2}}=3\] Since, \[\text{NaOH}\]is in excess. \[\therefore \]Remaining milliequivalents \[=[O{{H}^{-}}]\] \[=3-2=1\] Remaining concentration of \[[O{{H}^{-}}]\] \[=\frac{1}{500}=2\times {{10}^{-3}}\] \[\therefore \] \[[{{H}^{+}}]=\frac{10{{\,}^{-14}}}{2\times {{10}^{-3}}}\] \[=5\times {{10}^{-12}}\] \[pH=-\log [{{H}^{+}}]\] \[=-\log (5\times {{10}^{-12}})\] \[=11.3010\]
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