Magnus Ehingers undervisning

Allt du behöver för A i Biologi, Kemi, Bioteknik, Gymnasiearbete m.m.

Kemi 2

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Kapitel 3

kapitel 3Kompletta lösningar (med videor) till utvalda övningsuppgifter i kapitel 3 i Ehinger: Kemi 2 (NA förlag).

Hoppa direkt till uppgift: 3.14, 3.17, 3.28

3.14

[Video kommer inom kort.]

CH3CH2NH2 + H2O ⇌ CH3CH2NH\({\sf _3^+}\) + OH

  [CH3CH2NH2] [CH3CH2NH\({\sf _3^+}\)] [OH]  
f.r. \[0,750\] \[0\] \[0\] M
\[-x\] \[+x\] \[+x\] M
v.j. \[0,750-x\] \[x\] \[x\] M


\(5,6 \cdot 10^{-4} = \frac {x \cdot x}{0,750 - x} = \frac {x^2}{0750 - x}\)

pq-formeln ger:

\(x = 0,0202158\)

\([\text{OH}^-] = x\text{M} = 0,0202158\text{M} \approx 0,020\text{M}\)

3.17

[Video kommer inom kort.]

a.

\(K_{\text{b}} = \frac {K_{\text{w}}}{K_{\text{a}}} = \frac {1,0 \cdot 10^{-14}\text{M}^2}{6,2 \cdot 10^{-9}\text{M}} = 1,61290323 \cdot 10^{-6}\text{M} \approx 1,6 \cdot 10^{-6}\text{M}\)

b.

Vi kan skriva kodeinet K:

K + H2O ⇌ KH+ + OH

\(K_{\text{b}} = \frac {[\text{KH}^+][\text{OH}^-]}{[\text{K}]}\)

   [K]  [KH+] [OH]  
f.r. \[0,0020\] \[0\] \[0\] M
\[-x\] \[+x\] \[+x\] M
v.j. \[0,0020-x\] \[x\] \[x\] M

\(1,6... \cdot 10^{-6}\text{M} = \frac {x \cdot x}{0,750 - x} = \frac {x^2}{0,750-x}\)

pq-formeln ger:

\(x = 5,59955 \cdot 10^{-5} = [\text{OH}^-]\)

\(\text{pOH} = -\lg[\text{OH}^-] = -\lg (5,59955 \cdot 10^{-5}) = 4,25184687\)

\(\text{pH} = 14,00 - \text{pOH} = 14,00 - 4,25184687 = 9,74815313 \approx 9,75\)

3.28.

[Video kommer inom kort.]

a.

HC2O\({\sf _4^-}\) ⇌ H+ + C2O\({\sf _4^{2-}}\)

  [HC2O\({\sf _4^-}\)] [H+] [C2O\({\sf _4^{2-}}\)]  
f.r. \[0,100\] \[0\] \[0\] M
\[-x\] \[+x\] \[+x\] M
v.j. \[0,100-x\] \[x\] \[x\] M


\(K_{\text{a}} = \frac {[\text{H}^+][\text{C}_2\text{O}_4^{2-}]}{[\text{HC}_2\text{O}_4^-]}\)

\(5,1 \cdot 10^{-5} = \frac {x \cdot x}{0,100 - x} \approx \frac {x^2}{0,100}\)

\(x = \sqrt{0,100 \cdot 5,1 \cdot 10^{-5}} = 0,00225832\)

\(x \ll 0,10\), därför är det OK att förumma \(x\) bredvid \(0,100\). Vi får då:

\([\text{H}^+] = x\text{M} = 0,00225832\text{M}\)

\(\text{pH} = -\lg[\text{H}^+] = -\lg 0,00225832 = 2,64621491 \approx 2,65\)

b.

HSO\({\sf _4^-}\) ⇌ H+ + SO\({\sf _4^{2-}}\)

  [HSO\({\sf _4^-}\)] [H+] [SO\({\sf _4^{2-}}\)]  
f.r. \[0,100\] \[0\] \[0\] M
\[-x\] \[+x\] \[+x\] M
v.j. \[0,100-x\] \[x\] \[x\] M


\(K_{\text{a}} = \frac {[\text{H}^+][\text{SO}_4^{2-}]}{[\text{HSO}_4^-]}\)

\(10^{-2,00} = \frac {x \cdot x}{0,100 - x} = \frac {x^2}{0,100 - x}\)

pq-formeln ger:

\(x_1 = -0,0370156\); orimligt

\(x_2 = 0,0270156\)

\([\text{H}^+] = x\text{M} = 0,0270156\text{M}\)

\(\text{pH} = -\lg[\text{H}^+] = -\lg 0,0270156 = 1,56838538 \approx 1,57\)

c.

Al(H2O)\({\sf _6^{3+}}\) ⇌ H+ + Al(H2O)6(OH)2+

  [Al(H2O)\({\sf _6^{3+}}\)] [H+] [Al(H2O)6(OH)2+]  
f.r. \[0,100\] \[0\] \[0\] M
\[-x\] \[+x\] \[+x\] M
v.j. \[0,100-x\] \[x\] \[x\] M


\(K_{\text{a}} = \frac {[\text{H}^+][\text{Al(H}_2\text{O)}_5\text{(OH)}^{2+}]}{[\text{Al(H}_2\text{O)}_6^{3+}]}\)

\(10^{-5,00} = \frac {x \cdot x}{0,100 - x} \approx \frac {x^2}{0,100}\)

\(x = \sqrt{0,100 \cdot 10^{-5,00}} = 0,001\)

\(x \ll 0,10\), därför är det OK att förumma \(x\) bredvid \(0,100\). Vi får då:

\([\text{H}^+] = x\text{M} = 0,001\text{M}\)

\(\text{pH} = -\lg[\text{H}^+] = -\lg 0,001 = 3,00\)

 

 

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