Question

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

Answer 1

`pOH=1.60`

Explanation:

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`.

Also see this old problem here for more treatment.