What is the pH of a solution that is \(1.2 \times 10^{-8} \mathrm{M}\) in \(\mathrm{KOH}\) ? Does your answer make sense? What assumption do we normally make that is not valid in this case?

Since KOH is a strong base and dissociates fully in water, the concentration of hydroxide ions (OH-) is equal to the concentration of KOH in the solution. In this case, the concentration of hydroxide ions is \(1.2 \times 10^{-8} \:M \).

Using the formula for pOH, we have: \[ pOH = -\log_{10} [\mathrm{OH^-}] \] So, we get: \[ pOH = -\log_{10} (1.2 \times 10^{-8}) = 7.92 \]

The relationship between pH and pOH is given by the equation: \[ pH + pOH = 14 \] Using the pOH calculated in Step 2, we can now find the pH: \[ pH = 14 - pOH = 14 - 7.92 = 6.08 \] The pH of the solution containing KOH with a concentration of \(1.2 \times 10^{-8} \:M\) is 6.08.

Our calculated pH value seems a bit odd because it is smaller than 7, which is unusual for a basic solution. The explanation for this is that the slight acidity of the solution comes from the autodissociation of water. The reason why this error occurred is because we assumed that the amount of OH- ions came exclusively from the dissociation of KOH, while ignoring the autodissociation of water. Normally, the assumption is valid when dealing with strong acids or bases with relatively high concentrations, as the contribution of water's autodissociation becomes less significant. However, in this case, the concentration of KOH is extremely low, which makes the autodissociation of water more important and renders our assumption invalid.