A) \[-\log 2\]
B) \[-\log 0.2\]
C) 1.0
D) 2.0
Correct Answer: C
Solution :
N.eq. for \[HCl=\frac{0.4}{1000}\times 50=0.02\] N.eq. for \[NaOH=\frac{0.2}{1000}\times 50=0.1\] Now \[[O{{H}^{-}}]\] left \[=0.1-0.02\] \[[O{{H}^{-}}]=.08=8\times {{10}^{-2}}\]M \[pOH=-\log 8\times {{10}^{-2}}\]M ; \[pOH=1.0\]You need to login to perform this action.
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