Help me please?? Thanks!

1 Answer
Aug 22, 2017

Answer:

#[HO^-]=1.78xx10^-11*mol*L^-1#

Explanation:

In aqueous solution, water undergoes the famous autoprotolysis reaction:

#2H_2O(l) rightleftharpoonsH_3O^+ + HO^-#

And under standard condition of #298*K#, and near #1*atm#...

#K_w=[HO^-][H_3O^+]=10^-14#;

Taking #"log_10# of BOTH sides........

#log_10K_w=log_10[HO^-]+log_10[H_3O^+]#.....on rearrangment...

#underbrace(-log_10[HO^-])_(pH)underbrace(-log_10[H_3O^+])_(pH)=underbrace(-log_10(10^-14))_14#

And thus our working relationship.......which you will have to commit to memory:

#pH+pOH=14#

We have #pH=3.25#, thus #pOH=10.75#, and .........

#[HO^-]=10^(-10.75)=1.78xx10^-11*mol*L^-1#......

For the use of #pH# in buffer equations, see here.