a.
\[K_\mathrm{b} = \frac {\mathrm{[C_5H_5NH^+]}[\mathrm{OH^-}]}{\mathrm{[C_5H_5N]}}\]
b.
C5H5N(aq) + H2O ⇌ C5H5NH+(aq) + OH–(aq)
|
| [C5H5N] | [ C5H5NH+] | [OH–] |
|
| f.r. | \[0,20\] | \[0\] | \[0\] | M |
| ∆ | \[-x\] | \[+x\] | \[+x\] | M |
| v.j. | \[0,20 - x\] | \[x\] | \[x\] | M |
\[\begin{aligned}K_\mathrm{b} &= \frac {\mathrm{[C_5H_5NH^+]}[\mathrm{OH^-}]}{\mathrm{[C_5H_5N]}} \\ 1,8 \cdot 10^{-9} &= \frac {x \cdot x}{0,20 - x} ≈ \frac {x^2}{0,20} \\ x^2 &= 1,8 \cdot 10^{-9} \cdot 0,20 \\ x &= \sqrt{1,8 \cdot 10^{-9} \cdot 0,20} = 1,89736660 \cdot 10^{-5} = \mathrm{[OH^-]}\end{aligned}\]
\(x \ll 0,20\) ⇒ OK att försumma \(x\) bredvid \(0,20\).
\[\begin{aligned}\mathrm{pOH} &= -\lg \mathrm{[OH^-]} = -\lg(1,89736660 \cdot 10^{-5}) = 4,72184875 \\ \mathrm{pH} &= \mathrm{p}K_\mathrm{w} - \mathrm{pOH} = 14,00 - 4,72184875 = 9,27815125 ≈ 9,28\end{aligned}\]
