A) \[1\,M\text{-}KOH\]
B) \[10\,M\,{{H}_{2}}S{{O}_{4}}\]
C) Chlorine-water
D) Water containing carbon-dioxide
Correct Answer: A
Solution :
The pH of 1 M-KOH will be the highest \[\text{Ist pH}=14-\text{pOH}\] \[=14-\text{log }\!\![\!\!\text{ O}{{\text{H}}^{-}}]\] \[=14+\text{log }1\] \[=14+0=14\]
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