Imibuzo Yezibonelo Ekhuluma Ngamandla E-Acids Nezisekelo
Amandla e-asidi kanye nesisekelo kuyimibono ebalulekile kumakhemikhali ekhombisa ukuthi i-asidi noma i-molecule yesisekelo inganikela kanjani nge-proton (H⁺) noma yamukele i-proton, ngokulandelana. Kule ngxoxo, sizobheka izibonelo eziningana zezinkinga ezizosisiza siqonde amandla e-asidi kanye nesisekelo, kanye nendlela yokubala i-pH kanye ne-pOH yazo. Ngaphambi kokuba singene ezingxoxweni zezinkinga, ake siqale siqonde izisekelo ze-asidi kanye nesisekelo.
Isingeniso kuMandla e-Acids kanye nezisekelo
Amandla e-asidi noma isisekelo anganqunywa ngezinga layo le-ionization esixazululweni. I-asidi enamandla yileyo cishe exutshwe ngokuphelele ne-ionization esixazululweni, kuyilapho i-asidi ebuthakathaka ixutshwe kancane ne-ionization. Kunjalo nangezisekelo.
Ama-Acid Aqinile Nama-Acid Abuthakathaka
Ama-asidi aqinile, njenge:
– I-Hydrochloric acid (HCl)
– I-asidi ye-sulfuric (H₂SO₄)
– I-asidi ye-nitriki (i-HNO₃)
Cishe i-ionized ngokuphelele esixazululweni. Isibonelo, i-HCl emanzini izohlukana ibe yi-H⁺ kanye ne-Cl⁻ cishe ngokuphelele.
Ama-asidi abuthakathaka, njenge:
– I-Acetic acid (CH₃COOH)
– I-asidi ye-hydrofluoric (HF)
– I-asidi ye-citric (C₆H₈O₇)
kuhlanganiswe kancane kuphela esixazululweni. Lokhu kusho ukuthi nganoma yisiphi isikhathi, amanye ama-molecule e-asidi ahlala esesimweni sama-molecule futhi awahlukani abe ama-ion.
Izisekelo Eziqinile Nezisekelo Ezibuthakathaka
Izisekelo eziqinile, njenge:
– I-Sodium hydroxide (NaOH)
– I-Potassium hydroxide (KOH)
– I-calcium hydroxide (Ca(OH)₂)
Cishe i-ionized ngokuphelele kwisisombululo samanzi, ikhiqiza inani elikhulu lama-OH⁻ ions.
Izisekelo ezibuthakathaka, njenge:
– I-Amoniya (NH₃)
– I-Methylamine (CH₃NH₂)
i-ionization eyingxenye kuphela esixazululweni, ikhiqiza amanani amancane kakhulu ama-OH⁻ ions.
Imibuzo Eyisibonelo Nengxoxo
Ake sizame ezinye zezibonelo zezinkinga ukuze siqonde kangcono lo mqondo.
Isibonelo Umbuzo 1: Ukubala i-pH ye-Acid Eqinile
Umbuzo:
Bala i-pH yesisombululo se-HCl esingu-0,1 M.
Ingxoxo:
I-HCl iyi-asidi enamandla, ngakho-ke izoyijikisa ngokuphelele esixazululweni:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
Njengoba i-HCl i-ionized ngokugcwele, ukuhlushwa kwe-[H⁺] kulingana nokuhlushwa kwe-HCl:
\[ [\umbhalo{H}^+] = 0,1 \, \umbhalo{M} \]
\[ \umbhalo{pH} = -\umbhalo{H}^+] \]
\[ \umbhalo{pH} = -\log(0,1) \]
\[ \umbhalo{pH} = 1 \]
Ngakho-ke, i-pH yesisombululo se-HCl esingu-0,1 M ingu-1.
Isibonelo Umbuzo 2: Ukubala i-pH ye-Acid Ebuthakathaka
Umbuzo:
Bala i-pH yesisombululo se-CH₃COOH esingu-0,1 M. Kuyaziwa ukuthi i-Ka (i-acid dissociation constant) ye-CH₃COOH ingu-\(1,8 \times 10^{-5}\).
Ingxoxo:
Kuma-asidi abuthakathaka, izinga le-ionization kumele licatshangelwe. I-equation yokulingana yokuhlukaniswa kwe-acetic acid yile:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{COO}^- \]
Ngokusebenzisa i-Ka:
\[ \text{Ka} = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \]
Ake sithi ukuhlukaniswa [H⁺] = [CH₃COO⁻] = x bese kuthi i-[CH₃COOH] yokuqala ibe ngu-0,1 M, bese kuthi:
\[ 1,8 \izikhathi 10^{-5} = \frac{x^2}{0,1 – x} \]
Njengoba u-Ka emncane, u-x anganakwa uma kuqhathaniswa no-0,1:
\[ 1,8 \izikhathi eziyi-10^{-5} \cishe \frac{x^2}{0,1} \]
\[ x^2 = 1,8 \izikhathi eziyi-10^{-6} \]
\[ x = \sqrt{1,8 \izikhathi eziyi-10^{-6}} \]
\[ x \cishe 1,34 \izikhathi eziyi-10^{-3} \]
Bese, [H⁺] ≈ \(1,34 \times 10^{-3} \):
\[ \umbhalo{pH} = -\umbhalo{H}^+] \]
\[ \text{pH} = -\log(1,34 \times 10^{-3}) \]
\[ \text{pH} \cishe 2,87 \]
Ngakho-ke, i-pH yesisombululo se-0,1 M CH₃COOH icishe ibe ngu-2,87.
Isibonelo Umbuzo 3: Ukubala i-pOH kanye ne-pH yezisekelo eziqinile
Umbuzo:
Bala i-pOH kanye ne-pH yesisombululo se-NaOH esingu-0,01 M.
Ingxoxo:
I-NaOH iyisisekelo esiqinile esifakwe i-ion ngokugcwele emanzini:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]
Njengoba i-NaOH i-ionized ngokugcwele, ukuhlushwa kwe-[OH⁻] kulingana nokuhlushwa kwe-NaOH:
\[ [\umbhalo{OH}^-] = 0,01 \, \umbhalo{M} \]
\[ \umbhalo{pOH} = -\umbhalo[\umbhalo{OH}^-] \]
\[ \umbhalo{pOH} = -\log(0,01) \]
\[ \umbhalo{pOH} = 2 \]
Kuyaziwa ukuthi i-pH ne-pOH zihlobene nge-equation:
\[ \text{pH} + \text{pOH} = 14 \]
Ngakho-ke,
\[ \umbhalo{pH} = 14 – \umbhalo{pOH} \]
\[ \umbhalo{pH} = 14 – 2 \]
\[ \umbhalo{pH} = 12 \]
Ngakho-ke, i-pOH yesisombululo se-NaOH esingu-0,01 M ingu-2, kanti i-pH ingu-12.
Isibonelo Umbuzo 4: Ukubala i-pH yesisekelo esibuthakathaka
Umbuzo:
Bala i-pH yesisombululo se-NH₃ esingu-0,1 M. Kuyaziwa ukuthi i-Kb (i-base dissociation constant) ye-NH₃ ingu-\(1,76 \times 10^{-5}\).
Ingxoxo:
I-NH₃ emanzini iyasabela ekukhiqizeni i-NH₄⁺ kanye ne-OH⁻:
\[ \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]
Ngokusebenzisa i-Kb,
\[ \text{Kb} = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \]
Ake sithi ukuhlukaniswa [OH⁻] = [NH₄⁺] = x, bese kuthi i-[NH₃] yokuqala ibe ngu-0,1 M, bese kuthi:
\[ 1,76 \izikhathi 10^{-5} = \frac{x^2}{0,1 – x} \]
Njengokucabanga, u-x mncane kakhulu uma kuqhathaniswa no-0,1:
\[ 1,76 \izikhathi eziyi-10^{-5} \cishe \frac{x^2}{0,1} \]
\[ x^2 = 1,76 \izikhathi eziyi-10^{-6} \]
\[ x = \sqrt{1,76 \izikhathi eziyi-10^{-6}} \]
\[ x \cishe 1,33 \izikhathi eziyi-10^{-3} \]
Bese, [OH⁻] ≈ \(1,33 \times 10^{-3}\):
\[ \umbhalo{pOH} = -\umbhalo[\umbhalo{OH}^-] \]
\[ \text{pOH} = -\log(1,33 \times 10^{-3}) \]
\[ \text{pOH} \cishe 2,88 \]
Ngemuva kwalokho, i-pH ibalwa kanje:
\[ \umbhalo{pH} = 14 – \umbhalo{pOH} \]
\[ \umbhalo{pH} = 14 – 2,88 \]
\[ \text{pH} \cishe 11,12 \]
Ngakho-ke, i-pH yesisombululo se-NH₃ esingu-0,1 M cishe ingama-11,12.
Isiphetho
Ukuqonda amandla ama-asidi nezisekelo kubalulekile kumakhemikhali. Ngokuqonda ukuthi ungabala kanjani i-pH ne-pOH, kanye nokwazi ukuthi ungawabona kanjani ama-asidi nezisekelo ezibuthakathaka nezinamandla, uzobe ulungele kangcono ukubhekana nezinkinga ezahlukahlukene zekhemistri ezihilela izixazululo. Izibonelo ezinikeziwe ngethemba lokuthi zizocacisa le mibono futhi zinikeze isisekelo esiqinile sokutadisha okwengeziwe.
