What is #pOH# of a solution prepared by adding a #15.8mg# mass of potassium hydroxide to #11mL# of solution?

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Answer 1 · 2017-05-09T23:00:03

#pOH=1.60#

Now..........#pOH=-log_10[HO^-]#...........

And we have #15.8*mg# #KOH# dissolved in #11*mL# of solution.

#[HO^-]=((15.8xx10^-3*g)/(56.11*g*mol^-1))/(11*mLxx10^-3*mL*L^-1)#

#=2.56xx10^-2*mol*L^-1#.

But by definition, #pOH=-log_10(2.56xx10^-2)=1.60#

What is #pH# of this solution? You should be able to answer immediately. Why? Because in aqueous solution, #pH+pOH=14#.

What is #pOH# of a solution prepared by adding a #15.8*mg# mass of potassium hydroxide to #11*mL# of solution?