a.
CO\(_3^{2-}\) + H2O ⇌ HCO\(_3^-\) + OH–
\[K_\mathrm{b, CO_3^{2-}} = \frac {[\mathrm{HCO_3^-][\mathrm{OH^-}]}}{[\mathrm{CO_3^{2-}}]}\]
| [HCO\(_3^-\)] | [OH–] | [CO\(_3^{2-}\)] | ||
| f.r. | \[0,100\] | \[0\] | \[0\] | M |
| ∆ | \[-x\] | \[+x\] | \[+x\] | M |
| v.j. | \[0,100-x\] | \[x\] | \[x\] | M |
\[\begin{aligned}2,1 \cdot 10^{-4} &= \frac {x \cdot x}{0,100-x} \\ 2,1 \cdot 10^{-4}(0,100-x) &= x^2 \\ 2,1 \cdot 10^{-5} - 2,1 \cdot 10^{-4}x &= x^2 \\ 0 &= x^2 + 2,1 \cdot 10^{-4}x - 2,1 \cdot 10^{-5}\end{aligned}\]
pq-formeln ger:
\[\begin{aligned} x &= 0,00447878 = [\mathrm{OH^-}] \\ \mathrm{pOH} &= -\log[\mathrm{OH^-}] = -\log 0,00447878 = 2,34884027 \\ \mathrm{pH} &= \mathrm{p}K_\mathrm{w} - \mathrm{pOH} = 14,00 - 2,34884027 = 11,6511597 ≈ 11,65\end{aligned}\]
b.
S2– + H2O ⇌ HS– + OH–
\[K_\mathrm{b, S^{2-}} = \frac {[\mathrm{HS^-][\mathrm{OH^-}]}}{[\mathrm{S^{2-}}]}\]
| [HS–] | [OH–] | [S2–] | ||
| f.r. | \[0,100\] | \[0\] | \[0\] | M |
| ∆ | \[-x\] | \[+x\] | \[+x\] | M |
| v.j. | \[0,100-x\] | \[x\] | \[x\] | M |
\[\begin{aligned}10^{-1,08} &= \frac {x \cdot x}{0,100 - x} \\ 10^{-1,08}(0,100 - x) &= x^2 \\ 10^{-2,08} - 10^{-1,08}x &= x^2 \\ 0 &= x^2 + 10^{-1,08}x - 10^{-2,08}\end{aligned}\]
pq-formeln ger:
\[\begin{aligned} x &= 0,00447878 = [\mathrm{OH^-}] \\ \mathrm{pOH} &= -\log[\mathrm{OH^-}] = -\log 0,0586476 = 1,23174976 \\ \mathrm{pH} &= \mathrm{p}K_\mathrm{w} - \mathrm{pOH} = 14,00 - 1,23174976 = 12,7682502 ≈ 12,77\end{aligned}\]
c.
H2BO\(_3^-\) + H2O ⇌ H3BO3 + OH–
\[K_\mathrm{b, H_2BO_3^-} = \frac {[\mathrm{H_3BO_3][\mathrm{OH^-}]}}{[\mathrm{H_2BO_3^-}]}\]
| [H3BO3] | [OH–] | [H2BO\(_3^-\)] | ||
| f.r. | \[0,100\] | \[0\] | \[0\] | M |
| ∆ | \[-x\] | \[+x\] | \[+x\] | M |
| v.j. | \[0,100-x\] | \[x\] | \[x\] | M |
\[\begin{aligned}1,7 \cdot 10^{-5} &= \frac {x \cdot x}{0,100 - x} \\ 1,7 \cdot 10^{-5}(0,100 - x) &= x^2 \\ 1,7 \cdot 10^{-6} - 1,7 \cdot 10^{-5}x &= x^2 \\ 0 &= x^2 + 1,7 \cdot 10^{-5}x - 1,7 \cdot 10^{-6}\end{aligned}\]
pq-formeln ger:
\[\begin{aligned} x &= 0,00129537 = [\mathrm{OH^-}] \\ \mathrm{pOH} &= -\log[\mathrm{OH^-}] = -\log 0,00129537 = 2,88760617 \\ \mathrm{pH} &= \mathrm{p}K_\mathrm{w} - \mathrm{pOH} = 14,00 - 2,88760617 = 11,1123938 ≈ 11,11\end{aligned}\]
