A) 2.699
B) 13.30
C) 11.3010
D) 1.330
Correct Answer: C
Solution :
Given, volume of HCl solution = 200 mL pH of\[HCl\]solution =2.0 \[\therefore \]\[[{{H}^{+}}]\]in\[HCl\]solution\[=1\times {{10}^{-2}}\] \[\therefore \]Milliequivalents of\[HCl=200\times 1\times {{10}^{-2}}\] \[=2\] Similarly, volume of\[NaOH\]solution = 300 mL pH of \[NaOH=12\] \[pOH\]of \[NaOH=14-12=2\] \[[O{{H}^{-}}]\]in\[NaOH\]solution\[=1\times {{10}^{-2}}\] \[\therefore \]Milliequivalents of\[NaOH\] \[=300\times 1\times {{10}^{-2}}=3\] Since,\[NaOH\]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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