Magnus Ehingers undervisning

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

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

6.8

[Video kommer i augusti–september 2021.]

a.

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

b.

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

6.11.

HNO₃(aq) → H⁺(aq) + NO\({\sf _3^-}\)

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

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

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

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\({\sf _4^{2-}}\)(aq)

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

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

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

6.25.

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

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

6.26.

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄ + 2H₂O

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

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

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

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