A) 12
B) 10
C) 8
D) 6
Correct Answer: A
Solution :
Millimoles of\[HCl=40\times 0.10\text{ }M\] = 4 mmol Millimoles of\[NaOH=10\times 0.45\] =4.5 mmol Since,\[NaOH\]is in excess, millimoles of\[[O{{H}^{-}}]\]in mixture \[=(4.5-4)\text{ }mmol\] \[=0.5\text{ }mmol\] Concentration of\[O{{H}^{-}}\]in mixture\[=\frac{0.5}{40+10}\] \[=\frac{0.5}{50}=0.01\] \[pOH=-\log [O{{H}^{-}}]\] \[=-\log [1\times {{10}^{-2}}]\] \[=2\] \[pH+pOH=14\] \[pH=14-2=12\]You need to login to perform this action.
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