a.
CH3COO– + H2O ⇌ CH3COOH + OH–
b.
\[K_\mathrm{b} = \frac {[\mathrm{CH_3COOH}][\mathrm{OH^-}]}{[\mathrm{CH_3COOH^-}]}\]
c.
NaCH3COO(s) → Na+(aq) + CH3COO–(aq)
CH3COO– H2O ⇌ CH3COOH + OH–
| [CH3COO–] | [CH3COOH] | [OH–] | ||
| f.r. | \[0,100\] | \[0\] | \[0\] | M |
| ∆ | \[-x\] | \[+x\] | \[+x\] | M |
| v.j. | \[0,10-x\] | \[x\] | \[x\] | M |
\[\begin{aligned}K_\mathrm{b} &= \frac {[\mathrm{CH_3COOH}][\mathrm{OH^-}]}{[\mathrm{CH_3COOH^-}]} \\ 5,6 \cdot 10^{-10} &= \frac {x \cdot x}{0,10 - x} ≈ \frac {x^2}{0,10} \\ x^2 &= 5,6 \cdot 10^{-10} \cdot 0,10 \\ x &= \sqrt{5,6 \cdot 10^{-10} \cdot 0,10} = 7,48331477 \cdot 10^{-6} \\ [\mathrm{OH^-}] &= x\mathrm{M} ≈ 7,5\cdot 10^{-6}\mathrm{M}\end{aligned}\]
Eftersom \(x \ll 0,10\) är det OK att försumma \(x\) bredvid \(0,10\).
