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 300 mL of \[{{10}^{-2}}MNaOH=\frac{{{10}^{-2}}}{{{10}^{3}}}\times 300=3\times {{10}^{-3}}\] Moles of \[O{{H}^{-}}\] left in 500 mL of solution \[=(3\times {{10}^{-3}})-(2\times {{10}^{-3}})\] \[=1\times {{10}^{-3}}\] No. of \[O{{H}^{-}}\] moles in \[1000\,mL=2\times {{10}^{-3}}\] \[[O{{H}^{-}}]=2\times {{10}^{-3}}M\] \[pOH=2.693\] \[pH=14-2.698=11.3020\approx 11\]
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