Kemi Opgave 41

Beregning af pH i svag base.

$LaTex: \begin{eqnarray} K_b &=& \frac{[S]\cdot [OH^{-}]}{[B]}\\ [S] &=& [OH^{-}]\\ [B] &=& C_b-[OH^{-}]\\ [B] &\approx& C_b\\ K_b &\approx& \frac{[OH^{-}]^2}{C_b}\\ K_b\cdot C_b &\approx& [OH^{-}]^2\\ 2\cdot pOH &\approx& pK_b -log(C_b)\\ pOH &\approx& \frac{1}{2}\cdot (pK_b-log(C_b))\\ pH &\approx& 14-\frac{1}{2}\cdot (pK_b-log(C_b)) \end{eqnarray}$