What is #[HO^-]# of a solution for which #pH=3.10#?

1 Answer
anor277
Mar 8, 2017

#[HO^-]=1.27xx10^-11*mol*L^-1#.............

Explanation:

We use the relationship #14=pH+pOH#

Now #pH=-log_10(7.90xx10^-4)=-(-3.10)=3.10#

And #pOH=14-3.10=10.80#

And we can now take antilogarithms,

#[HO^-]=10^(-10.8)=1.27xx10^-11*mol*L^-1#

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