1414 Sýru-basa jafnvægi

Lykiljöfnur

K_{w} = [{H_{3} O}^{+}] [{O H}^{-}] = 1,0 \times 10^{- 14} (við 25 °C)

pH = - \log [{H_{3} O}^{+}]

pOH = - \log [{O H}^{-}]

[{H_{3} O}^{+}] = 10^{- pH}

[{O H}^{-}] = 10^{- pOH}

pH + pOH = {pK}_{w} = 14,00 við 25 °C

K_{a} = \frac{[{H_{3} O}^{+}] [A^{-}]}{[H A]}

K_{b} = \frac{[{H B}^{+}] [{O H}^{-}]}{[B]}

K_{a} \times K_{b} = 1,0 \times 10^{- 14} = K_{w}

prósentujónun = \frac{{[{H_{3} O}^{+}]}_{jafnv.}}{{[H A]}_{0}} \times 100

{pK}_{a} = - \log K_{a}

{pK}_{b} = - \log K_{b}

pH = {pK}_{a} + \log \frac{[A^{-}]}{[H A]}