What is the resultant #pOH# when a #5*mL# volume of #NaOH# of #0.165*mol*L^-1# concentration is diluted to a #450*mL# volume?

1 Answer
May 13, 2017

Answer:

#pOH=2.74#

Explanation:

#pOH=-log_(10)[HO^-]#; i.e. it is the basic counterpart to the #pH# scale.

And so we find #[HO^-]# of the given solution...........

#=(5xx10^-3*Lxx0.165*mol*L^-1)/(0.450*L)=1.83xx10^-3*mol*L^-1.#

And so #pOH=-log_10(1.83xx10^-3)=2.74#.

And further we know that #pH+pOH=14# for aqueous solution......You should be able to tell #pH# of this solution #"pdq"#. What is #pH# here?