What is the #pH# of an aqueous solution in which #[HO^-]=0.041*mol*L^-1#?

1 Answer
May 10, 2017

Answer:

#pH=12.6#

Explanation:

We know that in aqueous solution under standard conditions.....

#pH+pOH=14#.

And #pOH=-log_(10)[HO^-]#; so here #pOH=-log_10(0.041)=#

#-log_(10)0.041=-(-1.39)=1.39#.

And thus #pH=14-pOH=12.6#

So is this solution acid or alkaline?