Magnus Ehingers undervisning

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a. 

HC2O\(_4^-\)  H+ + C2O\(_4^{2-}\)

\[K_\mathrm{a, HC_2O_4^-} = \frac {\mathrm{[H^+][C_2O_4^{2-}]}}{\mathrm{[HC_2O_4^-}]}\]

 

[HC2O\(_4^-\)]

[H+]

[C2O\(_4^{2-}\)]

 

f.r.

\[0,100\]

\[0\]

\[0\]

M

\[-x\]

\[+x\]

\[+x\]

M

v.j.

\[0,100-x\]

\[x\]

\[x\]

M

\[5,1 \cdot 10^{-5} = \frac {x \cdot x}{0,100 - x}\]

\(pq\)-formeln ger:

\[\begin{aligned}x &= 0,00223296 = \mathrm{[H^+]} \\ \mathrm{pH} &= -\lg[\mathrm{H^+}] = -\lg 0,00223296 = 2,65111906 ≈ 2,65\end{aligned}\]

b. 

HSO\(_4^-\)  H+ + SO\(_4^-\)

\[K_\mathrm{a, HSO_4^-} = \frac {\mathrm{[H^+][SO_4^{2-}]}}{\mathrm{[HSO_4^-]}}\]

 

[HSO\(_4^-\)]

[H+]

[SO\(_4^-\)]

 

f.r.

\[0,100\]

\[0\]

\[0\]

M

\[-x\]

\[+x\]

\[+x\]

M

v.j.

\[0,100-x\]

\[x\]

\[x\]

M

\[\begin{aligned}K_\mathrm{a, HSO_4^-} &= 10^{-\mathrm{p}K_\mathrm{a, HSO_4^-}}\mathrm{M} = 10^{-2,00}\mathrm{M} \\ 10^{-2,00} &= \frac {x \cdot x}{0,100 - x}\end{aligned}\]

\(pq\)-formeln ger:

\[\begin{aligned}x &= 0,0270156 = \mathrm{[H^+]} \\ \mathrm{pH} &= -\lg[\mathrm{H^+}] = -\lg 0,0270156 = 1,56838538 ≈ 1,57\end{aligned}\]

c. 

Al(H2O)\(_6^{3+}\)  H+ + Al(OH)(H2O)\(_5^{2+}\)

\[K_\mathrm{a, Al(H_2O)_6^{3+}} = \frac {\mathrm{[H^+][Al(OH)(H_2O)_6^{2+}]}}{[\mathrm{Al(H_2O)_6^{3+}}]}\]

 

[Al(H2O)\(_6^{3+}\)]

[H+]

[Al(OH)(H2O)\(_5^{2+}\)]

 

f.r.

\[0,100\]

\[0\]

\[0\]

M

\[-x\]

\[+x\]

\[+x\]

M

v.j.

\[0,100-x\]

\[x\]

\[x\]

M

\[\begin{aligned}K_\mathrm{a, Al(H_2O)_6^{3+}} &= 10^{-\mathrm{p}K_\mathrm{a, Al(H_2O)_6^{3+}}}\mathrm{M} = 10^{-5,00}\mathrm{M} \\ 10^{-5,00} &= \frac {x \cdot x}{0,100 - x} ≈ \frac {x^2}{0,100} \\ 10^{-5,00} \cdot 0,100 &= x^2 \\ x &= \sqrt{10^{-5,00} \cdot 0,100} = 0,001 = \mathrm{[H^+]} \\ \mathrm{pH} &= -\lg [\mathrm{H^+}] = -\lg 0,001 = 3,00\end{aligned}\]