Lösningar

  1. \(c_1V_1 = c_2V_2\)
    \(c_2 = \frac {c_1V_1}{V_2} = \frac {0,200\mathrm{mol/dm^3} \cdot 50,0\mathrm{cm^3}}{750,0\mathrm{cm^3}} = 0,01333333\mathrm{mol/dm^3} \hspace{100cm}\)
    \(\mathrm{pOH} = -\lg[\mathrm{OH^-}] = -\lg0,01333333 = 1,8750613 \hspace{100cm}\)
    \(\mathrm{pH} = 14,00 - \mathrm{pOH} = 14,00 - 1,8750613 = 12,1249387 ≈ 12,12 \hspace{100cm}\)

  2. NaOH(aq) + HCl(aq) → H2O + NaCl(aq)
    \(n_\mathrm{NaOH} = \frac {m_\mathrm{NaOH}}{m_\mathrm{NaOH}} = \frac {450\mathrm{g}}{(23,0+16,0+1,01\mathrm{g/mol}} = 11,2471882\mathrm{mol}\)
    Reaktionsformeln ger att \(n_\mathrm{HCl} = n_\mathrm{NaOH} = 11,2471882\mathrm{mol}\).
    \(V_\mathrm{HCl} = \frac {n_\mathrm{HCl}}{c_\mathrm{HCl}} = \frac {11,2471882\mathrm{mol}}{1,0\mathrm{mol/dm^3}} = 11,2471882\mathrm{dm^3} \approx 11\mathrm{dm^3}\)

  3. KOH(aq) + HCl(aq) → KCl(aq) + H2O
    \(n_\mathrm{HCl} = c_\mathrm{HCl} \cdot V_\mathrm{HCl} = 0,100\mathrm{mol/dm^3} \cdot 37,5 \cdot 10^{-3}\mathrm{dm^3} = 0,00375\mathrm{mol}\)
    Reaktionsformeln ger att \(n_\mathrm{KOH} = n_\mathrm{HCl} = 0,00375\mathrm{mol}\).
    \(c_\mathrm{KOH} = \frac {n_\mathrm{KOH}}{V_\mathrm{KOH}} = \frac {0,00375\mathrm{mol}}{25,0 \cdot 10^{-3}\mathrm{dm^3}} = 0,150\mathrm{mol/dm^3}\)

  4. 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O
    \(n_\mathrm{NaOH} = c_\mathrm{NaOH} \cdot V_\mathrm{NaOH} = 2,00\mathrm{mol/dm^3} \cdot 20,0 \cdot 10^{-3}\mathrm{dm^3} = 0,0400\mathrm{mol}\)
    Reaktionsformeln ger att \(n_\mathrm{H_2SO_4} = \frac {1}{2}n_\mathrm{NaOH} = \frac {1}{2} \cdot 0,0400\mathrm{mol} = 0,0200\mathrm{mol}\).
    \(V_\mathrm{H_2SO_4} = \frac {n_\mathrm{H_2SO_4}}{c_\mathrm{H_2SO_4}} = \frac {0,0200\mathrm{mol}}{4,00\mathrm{mol/dm^3}} =  0,00500\mathrm{dm^3} = 5,00\mathrm{ml}\)

  5. Vid upplösning av de båda jonföreningarna bildas vattenlösta hydroxidjoner enligt nedanstående:
    Ba(OH)2(s) → Ba2+(aq) + 2OH(aq)
    KOH(s) → K+(aq) + OH(aq)

    Den totala substansmängden hydroxidjoner beräknas:
    \(n_\mathrm{OH^-} = 2n_\mathrm{Ba(OH)_2} + n_\mathrm{KOH} = 2 \cdot 0,2\mathrm{mol} + 0,4\mathrm{mol} = 0,8\mathrm{mol}\)

    Vid neutralisationen sker följande reaktion:
    H2SO4(aq) + 2OH(aq) → 2H2O + SO\({\sf _4^{2-}}\)

    Reaktionsformeln ger att \(n_\mathrm{H_2SO_4} = \frac {1}{2}n_\mathrm{OH^-} = \frac {1}{2} \cdot 0,8\mathrm{mol} = 0,4\mathrm{mol}\).