Magnus Ehingers undervisning

— Allt du behöver för A i Biologi, Kemi, Bioteknik, Gymnasiearbete med mera.

Under sommaren 2025 bygger jag om mina sidor för Biologi 1 och Kemi 1 inför Gy2025. 🏗 ⚒️ Snart är det klart! (Biologi 2 och Kemi 2 byggs om till Gy2025 under sommaren 2026.)

Ser du något som inte ser ut som det ska? Konstiga tecken, döda länkar och sådant? Mejla mig gärna och berätta, så ska jag försöka rätta till det så snart som möjligt!

Kompletta lösningar (med videor) till utvalda övningsuppgifter i kapitel 6 i Ehinger: Kemi 1 (NA förlag).

Hoppa direkt till uppgift: 6.8, 6.9, 6.11, 6.15, 6.16, 6.18, 6.25, 6.26

Kapitel 6

6.8

a.

\[\text{pH} = -\lg[\mathrm{H_3O^+]} = -\lg{0,0032} = 2,49485002 \approx 2,49 \hspace{100cm}\]

b.

\[\text{pH} = -\lg[\mathrm{H_3O^+]} = -\lg(6,3\cdot 10^{-9}) = 8,20065945 \approx 8,20 \hspace{100cm}\]

6.9.

a.

\[[\mathrm{H_3O^+}] = 10^{-\mathrm{pH}}\mathrm{M} = 10^{-1,25}\mathrm{M} = 0,05623413\mathrm{M} \approx 0,056\mathrm{M}\hspace{100cm}\]

b.

\[[\mathrm{H_3O^+}] = 10^{-\mathrm{pH}}\mathrm{M} = 10^{-9,40}\mathrm{M} = 3,98107171 \cdot 10^{-10}\mathrm{M} \approx 4,0 \cdot 10^{-10}\mathrm{M}\hspace{100cm}\]

6.11.

HNO3(aq) → H+(aq) + NO\(_3^-\)

\[n_{\text{HNO}_3}:n_{\text{H}^+} = 1:1 \Rightarrow n_{\text{H}^+} = n_{\text{HNO}_3} \hspace{100cm}\]

\[[\text{H}^+] = c_{\text{HNO}_3} = 0,0125\text{M} \hspace{100cm}\]

\[\text{pH} = -\lg[\text{H}^+] = -\lg 0,0125 = 1,90308999 \approx 1,903 \hspace{100cm}\]

6.15.

HCOOH(aq) → H+(aq) + COO(aq)

\[n_{\text{HCOOH}}:n_{\text{H}^+} = 1:1 \Rightarrow n_{\text{H}^+} = 0,004 \cdot n_{\text{HCOOH}} \hspace{100cm}\]

\[[\text{H}^+] = 0,004 \cdot c_{\text{HCOOH}} = 0,004 \cdot 0,10\text{M} = 0,0004\text{M} \hspace{100cm}\]

\[\text{pH} = -\lg[\text{H}^+] = -\lg0,0004 = 3,39794009 \approx 3,40 \hspace{100cm}\]

 

6.16.

H2SO4(aq) → 2H+(aq) + SO\(_4^{2-}\)(aq)

\[n_{\text{H}_2\text{SO}_4}:n_{\text{H}^+} = 1:2 \Rightarrow n_{\text{H}^+} = 2n_{\text{H}_2\text{SO}_4} \hspace{100cm}\]

\[[\text{H}^+] = 2c_{\text{H}_2\text{SO}_4} = 2 \cdot 9,65 \cdot 10^{-5}\text{M} = 0,000193\text{M} \hspace{100cm}\]

\[\text{pH} = -\lg[\text{H}^+] = -\lg0,000193 = 3,71444269 \approx 3,71 \hspace{100cm}\]

6.18.

a.

HCl(aq) + NaOH(aq) → H2O + NaCl(aq)

  HCl + NaOH → H2O + NaCl(aq)
 \[V\]

\[25,0\mathrm{ml}\]

\[16,7\mathrm{ml}\]

 
 \[n\]

\[0,00250\mathrm{mol}\]

\[0,00250\mathrm{mol}\]

 

\[V = 25,0\mathrm{ml} = 0,0250\mathrm{dm^3} \hspace{100cm}\]

\[n_\mathrm{HCl} = c_\mathrm{HCl} \cdot V_\mathrm{HCl} = 0,100\mathrm{mol/dm^3} \cdot 0,0250\mathrm{dm^3} = 0,00250\mathrm{mol} \hspace{100cm}\]

\[n_\mathrm{NaOH}:n_\mathrm{HCl} = 1:1 \hspace{100cm}\]

\[n_\mathrm{NaOH} = n_\mathrm{HCl} = 0,00250\mathrm{mol} \hspace{100cm}\]

\[V_\mathrm{NaOH} = \frac {n_\mathrm{NaOH}}{c_\mathrm{NaOH}} = \frac {0,00250\mathrm{mol}}{0,150\mathrm{mol/dm^3}} = 0,01666667\mathrm{dm^3} \approx 16,7\mathrm{ml} \hspace{100cm}\]

b.

H2SO4(aq) + 2NaOH(aq) → 2H2O + Na2SO4(aq)

  H2SO4 + 2NaOH → 2H2O + Na2SO4(aq)
 \[V\]

\[25,0\mathrm{ml}\]

\[50,0\mathrm{ml}\]

 
 \[n\]

\[0,00375\mathrm{mol}\]

\[0,0075\mathrm{mol}\]

 

\[V = 25,0\mathrm{ml} = 0,0250\mathrm{dm^3} \hspace{100cm}\]

\[n_\mathrm{H_2SO_4} = c_\mathrm{H_2SO_4} \cdot V_\mathrm{H_2SO_4} = 0,150\mathrm{mol/dm^3} \cdot 0,0250\mathrm{dm^3} = 0,00375\mathrm{mol} \hspace{100cm}\]

\[n_\mathrm{NaOH}:n_\mathrm{H_2SO_4} = 1:2 \hspace{100cm}\]

\[n_\mathrm{NaOH} = 2n_\mathrm{H_2SO_4} = 2 \cdot 0,00375\mathrm{mol} = 0,0075\mathrm{mol} \hspace{100cm}\]

\[V_\mathrm{NaOH} = \frac {n_\mathrm{NaOH}}{c_\mathrm{NaOH}} = \frac {0,0075\mathrm{mol}}{0,150\mathrm{mol/dm^3}} = 0,0500\mathrm{dm^3} = 50,0\mathrm{ml} \hspace{100cm}\]

c.

H3PO4(aq) + 3NaOH(aq) → 3H2O + Na3PO4(aq)

  H3PO4 + 3NaOH → H2O + Na3PO4
 \[V\]

\[25,0\mathrm{ml}\]

\[100\mathrm{ml}\]

 
 \[n\]

\[0,00500\mathrm{mol}\]

\[0,0150\mathrm{mol}\]

 

\[V = 25,0\mathrm{ml} = 0,0250\mathrm{dm^3} \hspace{100cm}\]

\[n_\mathrm{H_3PO_4} = c_\mathrm{H_3PO_4} \cdot V_\mathrm{H_3PO_4} = 0,200\mathrm{mol/dm^3} \cdot 0,0250\mathrm{dm^3} = 0,00500\mathrm{mol} \hspace{100cm}\]

\[n_\mathrm{H_3PO_4}:n_\mathrm{NaOH} = 1:3 \hspace{100cm}\]

\[n_\mathrm{NaOH} = 3n_\mathrm{H_3PO_4} = 3 \cdot 0,00500\mathrm{mol} = 0,0150\mathrm{mol} \hspace{100cm}\]

\[V_\mathrm{NaOH} = \frac {n_\mathrm{NaOH}}{c_\mathrm{NaOH}} = \frac {0,0150\mathrm{mol}}{0,150\mathrm{mol/dm^3}} = 0,100\mathrm{dm^3} = 100\mathrm{ml} \hspace{100cm}\]

6.25.

\[n_{\text{HCl}} = n_{\text{NaOH}} = \frac {m_{\text{NaOH}}}{M_{\text{NaOH}}} = \frac {450\text{g}}{(23,0+16,0+1,01)\text{g/mol}} = 11,2471882\text{mol} \hspace{100cm}\]

\[V_{\text{HCl}} = \frac {n_{\text{HCl}}}{c_{\text{HCl}}} = \frac {11,2471882\text{mol}}{1,0\text{mol/dm}^3} = 11,2471882\text{dm}^3 \approx 11\text{dm}^3 \hspace{100cm}\]

6.26.

H2SO4(aq) + 2NaOH(aq) → Na2SO4 + 2H2O

\[n_{\text{NaOH}} = c_{\text{NaOH}} \cdot V_{\text{NaOH}} = 2,00\text{mol/dm}^3 \cdot 0,0200\text{dm}^3 = 0,0400\text{mol} \hspace{100cm}\]

\[n_{\text{H}_2\text{SO}_4}:n_{\text{NaOH}} = 1:2 \hspace{100cm}\]

\[n_{\text{H}_2\text{SO}_4} = \frac {1}{2}n_{\text{NaOH}} = \frac {1}{2} \cdot 0,0400\text{mol} = 0,0200\text{mol} \hspace{100cm}\]

\[V_{\text{H}_2\text{SO}_4} = \frac {n_{\text{H}_2\text{SO}_4}}{c_{\text{H}_2\text{SO}_4}} = \frac {0,0200\text{mol}}{4,00\text{mol/dm}^3} = 0,00500\text{dm}^3 = 5,00\text{ml} \hspace{100cm}\]