A) 1
B) 5
C) 8
D) 9
E) 13
Correct Answer: E
Solution :
Moles of \[HCl=0.365\] Moles of \[NaOH=100\times 0.2=20\] Moles of\[NaOH\]left\[=19.635=20\] Moles of\[NaOH\]per litre \[=\frac{20\times {{10}^{-3}}\times 1000}{1000}\] \[pOH=-\log (2\times {{10}^{-2}})\] \[=1.7000=1.000\] \[pOH=1\] \[pH=14-1=13\]You need to login to perform this action.
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