question_answer 76)

Assuming complete dissociation, calculate the pH of following solutions: (i) 0.003 M HCl (b) 0.005 M \[NaOH\](c) 0.002 M \[HBr\](d) 0.002 M KOH.

Answer:

(a) \[HCl\to {{H}^{+}}+C{{l}^{-}}\] \[[{{H}^{+}}]=[HCl]=0.003M\] \[pH=-\log [{{H}^{+}}]\] \[=-\log [0.003]=2.52\] (b) \[NaOH\to N{{a}^{-}}+O{{H}^{-}}\] \[[O{{H}^{-}}]=[NaOH]=0.005M\] \[pH=14-2.301=11.699\] (c) \[HBr\to {{H}^{+}}+B{{r}^{-}}\] \[[{{H}^{+}}]=[HBr]=0.002M\] \[pH=-\log [{{H}^{+}}]=-log[0.002]=2.6989\] (d) \[KOH\to {{K}^{+}}+O{{H}^{-}}\] \[[O{{H}^{-}}]=[KOH]=0.002M\] \[pOH=-\log [O{{H}^{-}}]\] \[=-\log [0.002]\] \[=2.6989\] \[pH=14-2.6989=11.3011\]