A) \[10\]
B) \[11\]
C) \[12\]
D) \[14\]
Correct Answer: B
Solution :
Moles of \[{{H}^{+}}\] in 200 mL of, \[{{10}^{-2}}MHCl=\frac{{{10}^{-2}}}{{{10}^{3}}}\times 200=2\times {{10}^{-3}}\] Moles of \[O{{H}^{-}}\] in \[300mL\]of, \[=(3\times {{10}^{-3}})-(2\times {{10}^{-3}})=1\times {{10}^{-3}}\] No. of \[O{{H}^{-}}\] moles in \[1000\text{ }mL\]\[=2\times {{10}^{-3}}\] \[[O{{H}^{-}}]=2\times {{10}^{-3}}M\] \[pOH=2.693\] \[pH=14-2.698=11.3020\approx 11\]
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