If #pH=0.325# for an aqueous solution, what is #[HO^-]#?

1 Answer
anor277
Jul 21, 2017

#[HO^-]~=3xx10^-14*mol*L^-1#. The low #pH# specifies AN ACIDIC solution....

Explanation:

#K_w=10^-14=[H_3O^+][HO^-]# under standard conditions of #298*K# and (near to) #1*atm# pressure.........

It is usually simpler to convert this #pH# and #pOH# values than pfaff about with exponential products........

#14=pH+pOH#; #pH=-log_10(0.325)=0.488#; and #pOH=13.51#..

#[HO^-]=10^(-13.51)*mol*L^-1=??*mol*L^-1#

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