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
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\({\sf _3^-}\)
\[n_{\text{HNO}_3}:n_{\text{H}^+} = 1:1 \Rightarrow n_{\text{H}^+} = n_{\text{HNO}_3} \hspace{100cm}\]
\[[\text{H}^+] = c_{\text{HNO}_3} = 0,0125\text{M} \hspace{100cm}\]
\[\text{pH} = -\lg[\text{H}^+] = -\lg 0,0125 = 1,90308999 \approx 1,903 \hspace{100cm}\]
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)
\[n_{\text{H}_2\text{SO}_4}:n_{\text{H}^+} = 1:2 \Rightarrow n_{\text{H}^+} = 2n_{\text{H}_2\text{SO}_4} \hspace{100cm}\]
\[[\text{H}^+] = 2c_{\text{H}_2\text{SO}_4} = 2 \cdot 9,65 \cdot 10^{-5}\text{M} = 0,000193\text{M} \hspace{100cm}\]
\[\text{pH} = -\lg[\text{H}^+] = -\lg0,000193 = 3,71444269 \approx 3,71 \hspace{100cm}\]
6.18.
a.
HCl(aq) + NaOH(aq) → H2O + NaCl(aq)
| HCl | + NaOH | → H2O + NaCl(aq) | |
| \[V\] |
① \[25,0\mathrm{ml}\] |
④ \[16,7\mathrm{ml}\] |
|
| \[n\] |
② \[0,00250\mathrm{mol}\] |
③ \[0,00250\mathrm{mol}\] |
①
\[V = 25,0\mathrm{ml} = 0,0250\mathrm{dm^3} \hspace{100cm}\]
②
\[n_\mathrm{HCl} = c_\mathrm{HCl} \cdot V_\mathrm{HCl} = 0,100\mathrm{mol/dm^3} \cdot 0,0250\mathrm{dm^3} = 0,00250\mathrm{mol} \hspace{100cm}\]
③
\[n_\mathrm{NaOH}:n_\mathrm{HCl} = 1:1 \hspace{100cm}\]
\[n_\mathrm{NaOH} = n_\mathrm{HCl} = 0,00250\mathrm{mol} \hspace{100cm}\]
④
\[V_\mathrm{NaOH} = \frac {n_\mathrm{NaOH}}{c_\mathrm{NaOH}} = \frac {0,00250\mathrm{mol}}{0,150\mathrm{mol/dm^3}} = 0,01666667\mathrm{dm^3} \approx 16,7\mathrm{ml} \hspace{100cm}\]
b.
H2SO4(aq) + 2NaOH(aq) → 2H2O + Na2SO4(aq)
| H2SO4 | + 2NaOH | → 2H2O + Na2SO4(aq) | |
| \[V\] |
① \[25,0\mathrm{ml}\] |
④ \[50,0\mathrm{ml}\] |
|
| \[n\] |
② \[0,00375\mathrm{mol}\] |
③ \[0,0075\mathrm{mol}\] |
①
\[V = 25,0\mathrm{ml} = 0,0250\mathrm{dm^3} \hspace{100cm}\]
②
\[n_\mathrm{H_2SO_4} = c_\mathrm{H_2SO_4} \cdot V_\mathrm{H_2SO_4} = 0,150\mathrm{mol/dm^3} \cdot 0,0250\mathrm{dm^3} = 0,00375\mathrm{mol} \hspace{100cm}\]
③
\[n_\mathrm{NaOH}:n_\mathrm{H_2SO_4} = 1:2 \hspace{100cm}\]
\[n_\mathrm{NaOH} = 2n_\mathrm{H_2SO_4} = 2 \cdot 0,00375\mathrm{mol} = 0,0075\mathrm{mol} \hspace{100cm}\]
④
\[V_\mathrm{NaOH} = \frac {n_\mathrm{NaOH}}{c_\mathrm{NaOH}} = \frac {0,0075\mathrm{mol}}{0,150\mathrm{mol/dm^3}} = 0,0500\mathrm{dm^3} = 50,0\mathrm{ml} \hspace{100cm}\]
c.
H3PO4(aq) + 3NaOH(aq) → 3H2O + Na3PO4(aq)
| H3PO4 | + 3NaOH | → H2O + Na3PO4 | |
| \[V\] |
① \[25,0\mathrm{ml}\] |
④ \[100\mathrm{ml}\] |
|
| \[n\] |
② \[0,00500\mathrm{mol}\] |
③ \[0,0150\mathrm{mol}\] |
①
\[V = 25,0\mathrm{ml} = 0,0250\mathrm{dm^3} \hspace{100cm}\]
②
\[n_\mathrm{H_3PO_4} = c_\mathrm{H_3PO_4} \cdot V_\mathrm{H_3PO_4} = 0,200\mathrm{mol/dm^3} \cdot 0,0250\mathrm{dm^3} = 0,00500\mathrm{mol} \hspace{100cm}\]
③
\[n_\mathrm{H_3PO_4}:n_\mathrm{NaOH} = 1:3 \hspace{100cm}\]
\[n_\mathrm{NaOH} = 3n_\mathrm{H_3PO_4} = 3 \cdot 0,00500\mathrm{mol} = 0,0150\mathrm{mol} \hspace{100cm}\]
④
\[V_\mathrm{NaOH} = \frac {n_\mathrm{NaOH}}{c_\mathrm{NaOH}} = \frac {0,0150\mathrm{mol}}{0,150\mathrm{mol/dm^3}} = 0,100\mathrm{dm^3} = 100\mathrm{ml} \hspace{100cm}\]
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
\[n_{\text{NaOH}} = c_{\text{NaOH}} \cdot V_{\text{NaOH}} = 2,00\text{mol/dm}^3 \cdot 0,0200\text{dm}^3 = 0,0400\text{mol} \hspace{100cm}\]
\[n_{\text{H}_2\text{SO}_4}:n_{\text{NaOH}} = 1:2 \hspace{100cm}\]
\[n_{\text{H}_2\text{SO}_4} = \frac {1}{2}n_{\text{NaOH}} = \frac {1}{2} \cdot 0,0400\text{mol} = 0,0200\text{mol} \hspace{100cm}\]
\[V_{\text{H}_2\text{SO}_4} = \frac {n_{\text{H}_2\text{SO}_4}}{c_{\text{H}_2\text{SO}_4}} = \frac {0,0200\text{mol}}{4,00\text{mol/dm}^3} = 0,00500\text{dm}^3 = 5,00\text{ml} \hspace{100cm}\]
