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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