Question

A beaker contains 100.0 mL of distilled water. We add 0.10 g of KOH. What is; a. the concentration of H+ ions? b. the pH of the solution?

Answer 1

`H^+=5.62xx10^-13*mol*L^-1`

Explanation:

Well, first we work out the concentration of `"potassium hydroxide."`

`[KOH]="moles of salt"/"volume of solution"`...and reasonably we assume that the solution volume is the SAME as the volume of the solvent used to prepare it IN THIS SCENARIO of low concentration.

`[KOH]=((0.10*g)/(56.11*g*mol^-1))/(100.0*mLxx10^-3*L*mL^-1)=0.0178*mol*L^-1`...

Now we were asked for the concentration of `[HO^-]` AND for `pH`... It is a fact that in aqueous solution under standard conditions...that `14=pH+pOH`

And `pOH=-log_10[HO^-]=-log_10(0.0178)=1.75`

`pH=14.0-1.75=12.25`...a basic solution as required...

`[H_3O^+]=10^(-12.25)*mol*L^-1=5.62xx10^-13*mol*L^-1`...