Imibuzo Yesibonelo kanye Nengxoxo Nge-Arrhenius Acids Nezisekelo
Ama-asidi nezisekelo kuyimibono eyisisekelo kumakhemikhali. Enye yemibono eyaziwa kakhulu yokuchaza ukuziphatha kwama-asidi nezisekelo yi-Arrhenius theory. Ngokusho kuka-Arrhenius, i-asidi iyinhlanganisela ethi, uma incibilikiswa emanzini, ikhiqize ama-ion e-H⁺ (noma i-H₃O⁺), kuyilapho isisekelo siyinhlanganisela ethi, uma incibilikiswa emanzini, ikhiqize ama-ion e-OH⁻. Kulesi sihloko, sizoxoxa ngezibonelo eziningana zezinkinga mayelana nama-asidi nezisekelo ngokusho kwe-Arrhenius theory, kanye nezixazululo zazo.
Isibonelo Umbuzo 1
Umbuzo:
Isixazululo se-hydrochloric acid (HCl) esingu-0,1 M sincibilikiswa emanzini. Bala ukuhlushwa kwama-ion e-H⁺ esixazululweni.
Ingxoxo:
Ngokusho kwe-Arrhenius theory, i-HCl iyi-asidi ngoba ikhiqiza ama-ion e-H⁺ uma incibilikiswa emanzini. I-equation ye-ionization ye-HCl emanzini imi kanje:
\[ \text{HCl} \kuya ku- \text{H}^+ + \text{Cl}^- \]
Ukuhlushwa kokuqala kwe-HCl kungu-0,1 M. Njengoba i-HCl iyi-asidi enamandla e-ionization ngokuphelele kusisombululo samanzi, i-molecule ngayinye ye-HCl izokhiqiza i-H⁺ ion eyodwa kanye ne-Cl⁻ ion eyodwa. Ngakho-ke, ukuhlushwa kwama-H⁺ ion kusisombululo nakho kulingana nokuhlushwa kwe-HCl, okungukuthi:
\[ [\umbhalo{H}^+] = 0,1 \umbhalo{M} \]
Ngakho-ke, ukuhlushwa kwama-ion e-H⁺ kusisombululo se-HCl esingu-0,1 M kungu-0,1 M.
Isibonelo Umbuzo 2
Umbuzo:
Bala i-pH yesisombululo se-nitric acid (HNO₃) esingu-0,01 M.
Ingxoxo:
I-HNO₃ iyi-asidi enamandla e-ionization ngokuphelele emanzini, ngakho-ke i-equation ye-ionization ye-HNO₃ imi kanje:
\[ \text{HNO}_3 \to \text{H}^+ + \text{NO}_3^- \]
Ukuhlushwa okunikeziwe kwe-HNO₃ kungu-0,01 M. Njengoba i-HNO₃ iyi-asidi enamandla, i-molecule ngayinye ye-HNO₃ izokhiqiza i-H⁺ ion eyodwa. Ngakho-ke, ukuhlushwa kwama-H⁺ ion esixazululweni kulingana nokuhlushwa kwe-HNO₃, okungukuthi:
\[ [\umbhalo{H}^+] = 0,01 \umbhalo{M} \]
I-pH ingabalwa kusetshenziswa ifomula:
\[ \umbhalo{pH} = -\umbhalo[\umbhalo{H}^+] \]
Ngokufaka ukuhlushwa kwama-ion e-H⁺ kufomula, sithola:
\[ \text{pH} = -\log (0,01) = -\log (10^{-2}) = 2 \]
Ngakho-ke, i-pH yesisombululo se-HNO₃ esingu-0,01 M ingu-2.
Isibonelo Umbuzo 3
Umbuzo:
Thola ukuhlushwa kwama-ion e-OH⁻ kusisombululo se-NaOH esingu-0,05 M.
Ingxoxo:
I-NaOH iyisisekelo esiqinile esishintsha ngokuphelele i-ionization emanzini, ngakho-ke i-equation ye-ionization ye-NaOH imi kanje:
\[ \umbhalo{NaOH} \kuya \umbhalo{Na}^+ + \umbhalo{OH}^- \]
Ukuhlushwa kwe-NaOH okunikeziwe kungu-0,05 M. Njengoba i-NaOH iyisisekelo esiqinile, i-molecule ngayinye ye-NaOH izokhiqiza i-OH⁻ ion eyodwa. Ngakho-ke, ukuhlushwa kwama-OH⁻ ion esixazululweni kulingana nokuhlushwa kwe-NaOH, okungukuthi:
\[ [\umbhalo{OH}^-] = 0,05 \umbhalo{M} \]
Ngakho-ke, ukuhlushwa kwama-ion e-OH⁻ kusisombululo se-NaOH esingu-0,05 M kungu-0,05 M.
Isibonelo Umbuzo 4
Umbuzo:
Bala i-pH yesisombululo se-0,01 M KOH.
Ingxoxo:
I-KOH iyisisekelo esiqinile esiba yi-ionization ngokuphelele emanzini. Isibalo se-ionization se-KOH emanzini simi kanje:
\[ \text{KOH} \kuya ku- \text{K}^+ + \text{OH}^- \]
Ukuhlushwa kwe-KOH okunikeziwe kungu-0,01 M. Njengoba i-KOH iyisisekelo esiqinile, i-molecule ngayinye ye-KOH izokhiqiza i-OH⁻ ion eyodwa. Ngakho-ke, ukuhlushwa kwama-OH⁻ ion esixazululweni kulingana nokuhlushwa kwe-KOH, okungukuthi:
\[ [\umbhalo{OH}^-] = 0,01 \umbhalo{M} \]
I-pH iyisilinganiso sokuhlushwa kwama-ion e-H⁺ esixazululweni. Singasebenzisa ubudlelwano phakathi kwama-ion e-pH, i-H⁺, nama-ion e-OH⁻ anikezwa umkhiqizo wama-ion wamanzi (Kw):
\[ \text{Kw} = [\text{H}^+][\text{OH}^-] = 1 \times 10^{-14} \]
Njengoba sazi inani le-[OH⁻], singabala i-[H⁺]:
\[ [\text{H}^+] = \frac{\text{Kw}}{[\text{OH}^-]} = \frac{1 \times 10^{-14}}{0,01} = 1 \times 10^{-12} \text{M} \]
I-pH ingabalwa kusetshenziswa ifomula:
\[ \umbhalo{pH} = -\umbhalo[\umbhalo{H}^+] \]
Ngokufaka ukuhlushwa kwama-ion e-H⁺ kufomula, sithola:
\[ \text{pH} = -\log (1 \times 10^{-12}) = 12 \]
Ngakho-ke, i-pH yesisombululo se-0,01 M KOH ingu-12.
Isibonelo Umbuzo 5
Umbuzo:
Isixazululo sine-pH = 3. Bala ukuhlushwa kwama-ion e-H⁺ esixazululweni.
Ingxoxo:
I-pH iyisilinganiso sokuhlushwa kwama-ion e-H⁺ esixazululweni futhi ingavezwa ngefomula:
\[ \umbhalo{pH} = -\umbhalo[\umbhalo{H}^+] \]
Ukuze sibale ukuhlushwa kwama-ion e-H⁺, sishintsha ifomula ibe yilokhu:
\[ [\text{H}^+] = 10^{-\text{pH}} \]
Ngokufaka amanani e-pH anikeziwe, sithola:
\[ [\umbhalo{H}^+] = 10^{-3} = 0,001 \umbhalo{M} \]
Ngakho-ke, ukuhlushwa kwama-ion e-H⁺ esixazululweni esine-pH = 3 kungu-0,001 M.
Isibonelo Umbuzo 6
Umbuzo:
Isixazululo se-NH₄OH sinokuhlushwa okungu-0,1 M. Bala i-pH yesisombululo uma inani le-Kb le-NH₄OH lingu-1,8 x 10⁻⁵.
Ingxoxo:
I-NH₄OH iyisisekelo esibuthakathaka esingayi-ioni ngokuphelele emanzini. Isibalo se-ionization se-NH₄OH emanzini simi kanje:
\[ \text{NH}_4\text{OH} \leftrightarrow \text{NH}_4^+ + \text{OH}^- \]
I-constant yokulingana ye-ionization ye-NH₄OH ibizwa ngokuthi i-Kb futhi inikezwa yi:
\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_4\text{OH}} \]
Njengoba i-NH₄OH iyisisekelo esibuthakathaka, sisebenzisa i-Kb ukubala ukuhlushwa kwama-ion angu-OH⁻. Uma sicabanga ukuthi i-\(x\) ukuhlushwa kwe-NH₄OH ene-ionized, i-equation ye-ionization iba:
\[ K_b = \frac{x^2}{0,1 – x} \cishe \frac{x^2}{0,1} \]
Njengoba inani le-Kb lincane kakhulu, sicabanga ukuthi \(0,1 – x \approx 0,1\):
\[ 1,8 \izikhathi eziyi-10^{-5} = \frac{x^2}{0,1} \]
\[ x^2 = 1,8 \izikhathi eziyi-10^{-6} \]
\[ x = \sqrt{1,8 \izikhathi eziyi-10^{-6}} \]
\[ x \cishe 1,34 \izikhathi eziyi-10^{-3} \umbhalo{ M} \]
Ngakho-ke, ukuhlushwa kwama-ion angu-OH⁻ kungu-\(1,34 \times 10^{-3} \text{ M}\).
Kusukela kulokhu kuhlushwa, singabala ukuhlushwa kwama-ion e-H⁺ sisebenzisa i-Kw:
\[ [\text{H}^+] = \frac{1 \times 10^{-14}}{1,34 \times 10^{-3}} \]
\[ [\text{H}^+] = 7,46 \times 10^{-12} \text{M} \]
i-pH:
\[ \text{pH} = -\log (7,46 \times 10^{-12}) \]
\[ \text{pH} \cishe 11,13 \]
Ngakho-ke, i-pH yesisombululo se-NH₄OH esingu-0,1 M cishe ingama-11,13.
Isiphetho
Ithiyori ye-Arrhenius inikeza isisekelo esibalulekile sokuqonda ukuthi ama-acid kanye nezisekelo ziziphatha kanjani esixazululweni. Kule ngxoxo, sibheke izibonelo eziningana zezinkinga ezibonisa ukuthi le mibono ingasetshenziswa kanjani ukubala amazinga e-H⁺ kanye ne-OH⁻ ions kanye ne-pH yezixazululo ezahlukahlukene. Ukuqonda le thiyori kubalulekile kunoma ngubani ophishekela umsebenzi wekhemistri, kungaba esikoleni samabanga aphezulu noma ezingeni lemfundo ephakeme.
