Imibuzo Eyisibonelo Ekhuluma Ngomkhiqizo Wokuncibilika Nokuncibilika
Ukuncibilika kanye nomkhiqizo wokuncibilika (Ksp) kuyimiqondo ebalulekile kumakhemikhali ahlobene nezixazululo ezigcwele kanye nokuncibilika kwento esixazululweni. Ukuqonda lesi sihloko kungasisiza ukubikezela ukuthi usawoti uzoncibilika kangakanani esixazululweni nokuthi izici ezahlukahlukene zithinta kanjani lokho kuncibilika. Lesi sihloko sizoxoxa ngezibonelo eziningana zezinkinga kanye nezixazululo zazo ezihlobene nokuncibilika kanye ne-Ksp.
Imiqondo Eyisisekelo
Ukuncibilika (S) inani eliphezulu lezinto ezingancibilika ku-solvent ukuze zakhe isixazululo esigcwele ekushiseni okunikeziwe. Ngokuvamile, ukuncibilika kuvezwa ngamayunithi e-molarity (mol/L).
Umkhiqizo Wokuncibilika (Ksp) uyinto engaguquki yokulingana yokuhlukaniswa kwama-electrolyte ancibilika kancane emanzini. I-Ksp inikeza ulwazi mayelana nokuthi usawoti uzoncibilika kangakanani emanzini futhi ungumkhiqizo wamazinga ama-ion esixazululweni esigcwele, ngalinye liphakanyiswe emandleni e-stoichiometric coefficient yalo.
Isibonelo, uma sine-AxBy yosawoti ehlukana ibe ama-ion ngokwe-equation:
\[ \umbhalo{AxBy (s)} \rightleftharpoons xA^{n+} (aq) + yB^{m-} (aq) \]
Ngakho-ke, i-Ksp ingabhalwa kanje:
\[ \text{Ksp} = [A^{n+}]^x [B^{m-}]^y \]
Imibuzo Nezingxoxo Eziyisibonelo
Isibonelo Umbuzo 1
Umbuzo:
Bala ukuncibilika kwe-\(AgCl\) (Ksp = \(1.8 \times 10^{-10}\)) emanzini ahlanzekile.
Ingxoxo:
I-Ksp \(AgCl\) = 1.8 x \(10^{-10}\) mol²/L²
Ukusabela kokuhlukana:
\[AgCl (s) \rightleftharpoons Ag^+ (aq) + Cl^- (aq)\]
Ake sithi ukuncibilika kwe-\(AgCl\) kungu-S mol/L. Khona-ke, ukugxilisa kwe-\(Ag^+\) kanye ne-\(Cl^-\) esixazululweni esigcwele kuzolingana ne-S mol/L.
Isibalo se-Ksp:
\[Ksp = [Ag^+][Cl^-]\]
Faka inani lika-S esikhundleni salo:
\[1.8 \izikhathi ezingu-10^{-10} = S \izikhathi ezingu-S\]
\[S^2 = 1.8 \izikhathi ezingu-10^{-10}\]
\[S = \sqrt{1.8 \izikhathi eziyi-10^{-10}} \]
\[S = 1.34 \izikhathi 10^{-5} \, \text{mol/L}\]
Ngakho-ke, ukuncibilika kwe-\(AgCl\) emanzini ahlanzekile kungu-\(1.34 \times 10^{-5}\) mol/L.
Isibonelo Umbuzo 2
Umbuzo:
Kuyini ukuncibilika kwe-\(CaF_2\) (Ksp = \(3.9 \times 10^{-11}\)) emanzini ahlanzekile?
Ingxoxo:
Ukusabela kokuhlukana:
\[CaF_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2F^- (aq)\]
Ake sithi ukuncibilika kwe-\(CaF_2\) kungu-S mol/L. Bese kuthi ukugxilisa kwe-\(Ca^{2+}\) esixazululweni esigcwele yi-S mol/L kanti ukugxilisa kwe-\(F^-\) kungu-2S mol/L.
Isibalo se-Ksp:
\[Ksp = [Ca^{2+}][F^-]^2\]
Faka inani lika-S esikhundleni salo:
\[3.9 \izikhathi 10^{-11} = S \izikhathi (2S)^2\]
\[3.9 \izikhathi 10^{-11} = S \izikhathi 4S^2\]
\[3.9 \izikhathi eziyi-10^{-11} = 4S^3\]
\[S^3 = \frac{3.9 \izikhathi eziyi-10^{-11}}{4}\]
\[S^3 = 9.75 \izikhathi ezingu-10^{-12}\]
\[S = \sqrt[3]{9.75 \izikhathi eziyi-10^{-12}}\]
\[S \cishe 2.1 \izikhathi eziyi-10^{-4} \, \text{mol/L}\]
Ngakho-ke, ukuncibilika kwe-\(CaF_2\) emanzini ahlanzekile kungu-\(2.1 \times 10^{-4}\) mol/L.
Isibonelo Umbuzo 3
Umbuzo:
Kuyini ukuncibilika kwe-\(PbCl_2\) (Ksp = \(1.7 \times 10^{-5}\)) kusisombululo se-\(0.1\) M \(HCl\)?
Ingxoxo:
Ukusabela kokuhlukana:
\[PbCl_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2Cl^- (aq)\]
Ake sithi ukuncibilika kwe-\(PbCl_2\) kungu-S mol/L. Bese kuthi ukugxilisa kwe-\(Pb^{2+}\) esixazululweni esigcwele yi-S mol/L kanti ukugxilisa okwengeziwe kwe-\(Cl^-\) kusukela ekuhlukanisweni kwe-\(PbCl_2\) kungu-2S mol/L.
Kodwa-ke, kukhona futhi i-\(Cl^-\) evela ku-\(HCl\) esevele ikhona kuze kufike ku-0.1 M.
Isibalo se-Ksp:
\[Ksp = [Pb^{2+}][Cl^-]^2\]
Faka amanani ka-S kanye no-(Cl^-\):
\[1.7 \izikhathi 10^{-5} = S \izikhathi (0.1 + 2S)^2\]
Njengoba u-0.1 M \(Cl^-\) mkhulu kakhulu kuno-2S, \(0.1 + 2S \cishe kube ngu-0.1\).
Ngakho-ke ukubala kuba lula:
\[1.7 \izikhathi 10^{-5} = S \izikhathi (0.1)^2\]
\[1.7 \izikhathi 10^{-5} = S \izikhathi 0.01\]
\[S = \frac{1.7 \times 10^{-5}}{0.01}\]
\[S = 1.7 \izikhathi 10^{-3} \, \text{mol/L}\]
Ngakho-ke, ukuncibilika kwe-\(PbCl_2\) ku-\(0.1\) M \(HCl\) kungu-\(1.7 \times 10^{-3}\) mol/L.
Isibonelo Umbuzo 4
Umbuzo:
Bala ukuncibilika kwe-\(BaSO_4\) (Ksp = \(1.1 \times 10^{-10}\)) kusisombululo esiqukethe \(0.01\) M \(Na_2SO_4\).
Ingxoxo:
Ukusabela kokuhlukana:
\[BaSO_4 (s) \rightleftharpoons Ba^{2+} (aq) + SO_4^{2-} (aq)\]
Ake sithi ukuncibilika kwe-\(BaSO_4\) kungu-S mol/L. Bese kuthi ukugxilisa kwe-\(Ba^{2+}\) esixazululweni esigcwele yi-S mol/L kanti ukugxilisa kwe-\(SO_4^{2-}\) kwe-\(BaSO_4\) kungu-S mol/L. Kodwa-ke, kukhona \(0.01\) M \(SO_4^{2-}\) kwe-\(Na_2SO_4\).
Isibalo se-Ksp:
\[Ksp = [Ba^{2+}][SO_4^{2-}]\]
Faka amanani ka-S kanye no-\(SO_4^{2-}\):
\[1.1 \izikhathi 10^{-10} = S \izikhathi (0.01 + S)\]
Njengoba i-\(0.01\) M \(SO_4^{2-}\) inkulu kakhulu kune-S, \(0.01 + S \cishe kube ngu-0.01\).
Ngakho-ke ukubala kuba lula:
\[1.1 \izikhathi 10^{-10} = S \izikhathi 0.01\]
\[S = \frac{1.1 \times 10^{-10}}{0.01}\]
\[S = 1.1 \izikhathi 10^{-8} \, \text{mol/L}\]
Ngakho-ke, ukuncibilika kwe-\(BaSO_4\) esixazululweni esiqukethe i-\(0.01\) M \(Na_2SO_4\) kungu-\(1.1 \times 10^{-8}\) mol/L.
Isiphetho
Ngezinkinga zesibonelo ezingenhla, singabona ukuthi imiqondo yokuncibilika kanye nomkhiqizo wokuncibilika (Ksp) isetshenziswa kanjani ezimweni ezahlukene. Lokhu kuqonda kubalulekile ekuhlaziyweni kwamakhemikhali, ikakhulukazi lapho sifuna ukunquma ukuncibilika kosawoti othize ngaphansi kwezimo ezahlukahlukene, njengasesinyibilikisini esimsulwa noma lapho kukhona i-ion evamile. Ikhono lokuxazulula lezi zinkinga lisisiza ezindleleni eziningi ezisebenzayo, okuhlanganisa imithi, ikhemistri yemvelo, kanye nokuhlanzwa kwezinto.
Kubalulekile ukuqaphela ukuthi ukuncibilika kukasawoti akuxhomekile kuphela enanini le-Ksp, kodwa futhi kungathinteka ukuhlushwa kwamanye ama-ion esixazululweni, izinga lokushisa, kanye ne-pH yendawo. Ngokusebenzisa izimiso eziyisisekelo zekhemistri kanye nokubala okulula, singabikezela futhi silawule izenzakalo zokuncibilika ezinhlelweni ezahlukene zamakhemikhali.
