What is pOHpOH of a solution prepared by adding a 15.8*mg15.8⋅mg mass of potassium hydroxide to 11*mL11⋅mL of solution?

1 Answer
anor277
May 9, 2017

pOH=1.60pOH=1.60

Explanation:

Now..........pOH=-log_10[HO^-]pOH=−log10[HO−]...........

And we have 15.8*mg15.8⋅mg KOHKOH dissolved in 11*mL11⋅mL of solution.

[HO^-]=((15.8xx10^-3*g)/(56.11*g*mol^-1))/(11*mLxx10^-3*mL*L^-1)[HO−]=15.8×10−3⋅g56.11⋅g⋅mol−111⋅mL×10−3⋅mL⋅L−1

=2.56xx10^-2*mol*L^-1=2.56×10−2⋅mol⋅L−1.

But by definition, pOH=-log_10(2.56xx10^-2)=1.60pOH=−log10(2.56×10−2)=1.60

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

Also see this [old problem here for more treatment.](https://socratic.org/questions/ph-value-definition)

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