Calculate the pH of a solution containing 1.0x10^{-8} M KOH |
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Source: Kipen Mangas
https://tapas.io/episode/1465471
WHEN WATER IS THE SOLVENT
When an acid or a base is added to water, the contribution of [H^{+}] or [OH^{-}] on the pH from the natural dissociation of water (hydrolysis or ionization) is very small and can often be neglected in most problems (99.9%). Therefore, the pH in water is mostly "driven" by the acid or base added. However, if the concentration of OH^{-} added (here KOH = 1.0x10^{-8} M) is close to the one of the water hydrolysis concentration, you need to do an ICE table to know the exact pH. ICE Table
The equilibrium runs backward since adding a product will shift the equilibrium to the left (leChatelier) K_{w} = 1.0x10^{-14} = (1.0x10^{-7} - x)(1.1x10^{-7} - x) 0 = x^{2} - 2.1x10^{-7}x + 1.0x10^{-15} Then, you need to solve this quadratic equation to get the answer. The solution is x = 4.88x10^{-9}, therefore, the [H^{+}] concentration is 1.00x10^{-7} - 4.88x10^{-9} = 9.51x10^{-8} pH = -log(9.51x10^{-8}) = 7.02. Note: This problem is similar to the one in Zumdahl 14-59c with HI and 14-89b with NaOH |