Question #4af57

1 Answer
Stefan V.
Jan 12, 2015

Sodium hydroxide is a strong base, which means it dissociates completely in aqueous solution into #"Na"^(+)# and #"OH"^(-)# ions.

This tells you that the concentrations of the #"Na"^(+)# cations and of the #"OH"^(-)# anions will be equal to each other, and to the initial concentration of #"NaOH"#.

Since you start with #10^(-5)# #"M"# #"NaOH"#, in aqueous solution you'll end up with

#["Na"^(+)] = 10^(-5)# #"M"# and #["OH"^(-)] = 10^(-5)# #"M"#

The equation to use in this case is

#pOH = -log([OH^(-)]) = -log(10^(-5)) = 5#

This implies a pH of

#pH = 14 - pOH = 14 - 5 =9#

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