A solution is prepared by adding of KOH to 1 liter of water. The resulting solution has an [OH negative]of 1 * 10 negative 3. What would be the pOH of this solution?
Can you show me how you came up with this answer? Thank you.
KOH is a strong base; it ionizes 100%.
KOH ==> K^+ + OH^-
The problem states that (OH^-) = 1E-3.
pOH = -log(OH^-)
pOH = -log(1E-3)
pOH = -(-3)
pOH = 3.