Question

solid potassium hydroxide is dissolved in pure water, increasing the ph of the solution to 12.12. what is the oh^{-} concentration of this solution?

Explanation:

Step1: Calculate pOH

We know that \(pH + pOH=14\). Given \(pH = 12.12\), then \(pOH=14 - pH\).
\(pOH=14 - 12.12=1.88\)

Step2: Calculate \([OH^{-}]\)

The formula for \(pOH\) is \(pOH=-\log[OH^{-}]\). Rearranging for \([OH^{-}]\), we get \([OH^{-}]=10^{-pOH}\).
Substituting \(pOH = 1.88\) into the formula, \([OH^{-}]=10^{- 1.88}\)
Using the property \(a^{-b}=\frac{1}{a^{b}}\), \(10^{-1.88}=\frac{1}{10^{1.88}}\approx0.0132\space mol/L\)

Answer:

The \(OH^{-}\) concentration of the solution is approximately \(0.0132\space mol/L\)