a.
\[\begin{aligned}\mathrm{p}K_\mathrm{w} &= \mathrm{pH + pOH} \\ \mathrm{pOH} &= \mathrm{p}K_\mathrm{w} - \mathrm{pH} = 14,00 - 11,18 = 2,83 \\ [\mathrm{OH^-}] &= 10^{-\mathrm{pOH}}\mathrm{M} = 10^{-2,83}\mathrm{M} = 0,00147911\mathrm{M} ≈ 0,0015\mathrm{M}\end{aligned}\]
b.
NH3 + H2O ⇌ NH\(_4^+\) + OH–
| [NH3] | [NH\(_4^+\)] | [OH–] | ||
| f.r. | \[0,125\] | \[0\] | \[0\] | M |
| ∆ | \[-10^{-2,83}\] | \[+10^{-2,83}\] | \[+10^{-2,83}\] | M |
| v.j. | \[0,125 - 10^{-2,83}\] | \[10^{-2,83}\] | \[10^{-2,83}\] | M |
\[\begin{aligned}K_\mathrm{b} &= \frac {[\mathrm{NH_4^+}][\mathrm{OH^-}]}{[\mathrm{NH_3}]} = \frac {10^{-2,83} \cdot 10^{-2,83}}{0,125 - 10^{-2,83}} = 1,77116729 \cdot 10^{-5}\mathrm{M} ≈ \\ &≈ 1,8 \cdot 10^{-5}\mathrm{M} \\ \mathrm{p}K_\mathrm{b} &= -\lg K_\mathrm{b} = -\lg(1,77116729 \cdot 10^{-5}) = 4,75174042 ≈ \\ &≈ 4,75\end{aligned}\]
