A) \[\frac{M}{10}HCl\]
B) \[\frac{M}{100}HCl\]
C) \[\frac{M}{10}NaOH\]
D) \[\frac{M}{100}NaOH\]
Correct Answer: C
Solution :
Key Idea: \[pH=-\log [{{H}^{+}}]\] \[pH=14-pOH\] Calculate pH of all the solution to find which will have maximum\[~pH.\] (a)\[M/10\,HCl\] \[\therefore \] \[[{{H}^{+}}]={{10}^{-1}}\] \[pH=-\log [{{H}^{+}}]\] \[=-\log [{{10}^{-1}}]\] \[=1\] (b) \[M/100\,HCl\] \[\therefore \] \[[{{H}^{+}}]={{10}^{-2}}\] \[pH=-\log [{{H}^{+}}]\] \[=-\log [{{10}^{-2}}]\] \[=2\] (c) \[M/10\,NaOH\] \[\therefore \] \[[O{{H}^{-}}]={{10}^{-1}}\] \[pH=14-pOH\] \[=14[-log{{10}^{-1}}]\] \[=14-1\] \[=13\] (d) \[M/100\,NaOH\] \[\therefore \] \[[O{{H}^{-}}]={{10}^{-2}}\] \[pH=14-pOH\] \[=14-[log{{10}^{-2}}]\] \[=14-2=12\] \[\therefore \]\[M/10\,NaOH\]solution has highest pH.
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