Question

the ph of a solution of koh is 11.36. what is the concentration of oh⁻ in this solution?

Explanation:

Step1: Calculate pOH

We know that \(pH + pOH=14\). Given \(pH = 11.36\), then \(pOH=14 - pH\).
\(pOH=14 - 11.36=2.64\)

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

The formula for \(pOH\) is \(pOH=-\log[OH^{-}]\). Rearranging for \([OH^{-}]\), we get \([OH^{-}]=10^{-pOH}\).
Substituting \(pOH = 2.64\) into the formula, \([OH^{-}]=10^{- 2.64}\)
Using the property \(a^{-b}=\frac{1}{a^{b}}\) and \(10^{m + n}=10^{m}\times10^{n}\), \(10^{-2.64}=\frac{1}{10^{2.64}}=\frac{1}{10^{2}\times10^{0.64}}\)
Since \(10^{0.64}\approx4.37\) (by calculator), \(\frac{1}{100\times4.37}\approx2.29\times10^{-3}\space M\)

Answer:

\(2.29\times 10^{-3}\space M\)