A) 10
B) 2.7
C) 4.0
D) 11.3
E) 2
Correct Answer: D
Solution :
Concentration of\[HCl\]solution\[=1\times {{10}^{-2}}\] \[\therefore \]Millimoles of\[HCl\]solution\[=200\times 1\times {{10}^{-2}}\] \[=2\] Similarly, millimoles of\[NaOH\]solution \[=300\times 1\times {{10}^{-2}}\] \[=3\] Concentration of the resultant solution \[=\frac{3-2}{300+200}\] \[=\frac{1}{500}=0.2\times {{10}^{-2}}\] \[\therefore \]\[[O{{H}^{-}}]=0.2\times {{10}^{-2}}\] \[pOH=-\log [O{{H}^{-}}]=-\log [2\times {{10}^{-3}}]=+2.70\] \[\therefore \]pH of the resulting mixture \[=14-2.70\] \[=11.3\]
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