Isibonelo Semibuzo Yengxoxo Ngama-Electrolyte
Ama-electrolyte yizinto ezingaqhuba ugesi uma zincibilikiswa emanzini noma kwezinye izinyibilikisi. Ama-electrolyte ahlukaniswe ngezinhlobo ezimbili eziyinhloko: ama-electrolyte aqinile nama-electrolyte abuthakathaka. Ama-electrolyte aqinile akhiqiza i-ion ngokuphelele esixazululweni, kuyilapho ama-electrolyte abuthakathaka enza i-ion kancane. Ama-electrolyte adlala indima ebalulekile ekusabeleni kwamakhemikhali ahlukahlukene kanye nokuphila kwansuku zonke. Kulesi sihloko, sizoxoxa ngezinkinga eziningana zezibonelo kanye nezincazelo zazo mayelana nama-electrolyte.
Isibonelo Umbuzo 1: Ukunquma Izinga Lokufakwa Kwe-Ionization
Umbuzo: Isixazululo se-Acetic acid (CH₃COOH) saziwa ukuthi sinokuhlushwa okungu-0,1 M kanye nezinga le-ionization (α) elingu-4%. Liyini izinga lama-ion esixazululweni?
Ingxoxo:
1. Thola izinga le-ionization:
Izinga le-ionization (α) liyisilinganiso sento exutshwe nge-ion kusisombululo. Njengoba i-α = 4% = 0.04.
2. I-equation ye-ionization ye-acetic acid:
\[
\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+
\]
3. Ukubala ukuhlushwa:
Ukuhlushwa kokuqala kwe-CH₃COOH kungu-0,1 M. Njengoba izinga le-ionization lingu-0,04, khona-ke:
\[
[\text{CH}_3\text{COO}^-] = [\text{H}^+] = 0.1 \times 0.04 = 0.004 \text{M}
\]
Ukuhlushwa kwe-CH₃COOH engaxutshwe ne-ionized:
\[
[\text{CH}_3\text{COOH}] = 0.1 \text{M} – 0.004 \text{M} = 0.096 \text{M}
\]
I-Jawaban:
\[
[\umbhalo{CH}_3\umbhalo{COO}^-] = 0.004 \umbhalo{M}
[\umbhalo{H}^+] = 0.004 \umbhalo{M}
[\text{CH}_3\text{COOH}] un-ionized = 0.096 \text{M}
\]
Isibonelo Umbuzo 2: Ukubala i-Ksp (Umkhiqizo Wokuncibilika)
Umbuzo: Uma unikezwe usawoti, i-BaSO₄, encibilika kancane emanzini enokuncibilika okungu-1,0 × 10⁻⁵ M. Bala i-Ksp ye-BaSO₄.
Ingxoxo:
1. Isibalo sokuncibilika kwe-BaSO₄ usawoti:
\[
\umbhalo{BaSO}_4 (s) \rightleftharpoons \umbhalo{Ba}^{2+} (aq) + \umbhalo{SO}_4^{2-} (aq)
\]
2. Ukuncibilika:
Njengoba kunikezwe ukuncibilika kwe-BaSO₄ = 1,0 × 10⁻⁵ M.
3. Bala ukuhlushwa kwama-ion:
Uma ukuncibilika kwe-BaSO₄ = s = 1,0 × 10⁻⁵ M, khona-ke:
\[
[\text{Ba}^{2+}] = 1,0 \times 10^{-5} \text{M}
\]
\[
[\text{SO}_4^{2-}] = 1,0 \times 10^{-5} \text{M}
\]
4. Ukubala i-Ksp:
\[
K_{sp} = [\text{Ba}^{2+}] \times [\text{SO}_4^{2-}]
\]
\[
K_{sp} = (1,0 \izikhathi 10^{-5}) \izikhathi (1,0 \izikhathi 10^{-5})
\]
\[
K_{sp} = 1,0 \izikhathi 10^{-10}
\]
I-Jawaban:
\[
K_{sp} \text{ BaSO₄} = 1,0 \times 10^{-10}
\]
Isibonelo Umbuzo 3: I-pH ye-Acid kanye ne-Base Solutions
Umbuzo: Bala i-pH yesisombululo se-HCl ngokuhlushwa okungu-0,01 M.
Ingxoxo:
1. I-equation ye-ionization ye-HCl:
\[
\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-
\]
I-HCl iyi-asidi enamandla, ngakho-ke i-ionization iqediwe.
2. Bala ukuhlushwa kwe-hydrogen ion:
Ukuhlushwa kwe-HCl = 0,01 M kusho:
\[
[\umbhalo{H}^+] = 0.01 \umbhalo{M}
\]
3. Ukubala i-pH:
\[
\text{pH} = -\log[\text{H}^+]
\]
\[
\umbhalo{pH} = -\log(0.01)
\]
\[
\umbhalo{pH} = 2
\]
I-Jawaban:
\[
\text{pH} \text{ HCl solution} = 2
\]
Isibonelo Umbuzo 4: Ukubala i-pOH yesisombululo esiyisisekelo
Umbuzo: Bala i-pOH kanye ne-pH yesisombululo se-KOH ngokuhlushwa okungu-0,001 M.
Ingxoxo:
1. I-equation ye-ionization ye-KOH:
\[
\text{KOH} \rightarrow \text{K}^+ + \text{OH}^-
\]
I-KOH iyisisekelo esiqinile, ngakho-ke i-ionization isiqediwe.
2. Bala ukuhlushwa kwama-ion e-hydroxide:
Ukuhlushwa kwe-KOH = 0,001 M kusho:
\[
[\umbhalo{OH}^-] = 0.001 \umbhalo{M}
\]
3. Ukubala i-pOH:
\[
\text{pOH} = -\log[\text{OH}^-]
\]
\[
\umbhalo{pOH} = -\log(0.001)
\]
\[
\umbhalo{pOH} = 3
\]
4. Ukubala i-pH:
\[
\umbhalo{pH} = 14 – \umbhalo{pOH}
\]
\[
\umbhalo{pH} = 14 – 3
\]
\[
\umbhalo{pH} = 11
\]
I-Jawaban:
\[
\text{pOH} \text{ isixazululo se-KOH} = 3
\text{pH} \text{ isixazululo se-KOH} = 11
\]
Isibonelo Umbuzo 5: Ukunquma Ukuhlushwa Kwama-Ion e-Hydroxide
Umbuzo: Isixazululo se-hydrofluoric acid (HF) sine-pH engu-3. Lingakanani izinga lama-ion e-hydroxide (OH⁻) esixazululweni?
Ingxoxo:
1. Thola ukuhlushwa kwama-ion e-hydrogen:
Njengoba i-pH = 3 inikezwe, khona-ke:
\[
[\umbhalo{H}^+] = 10^{-3} \umbhalo{M}
\]
2. Ukusebenzisa umqondo wobudlelwano phakathi kwe-pH ne-pOH:
\[
\umbhalo{pH} + \umbhalo{pOH} = 14
\]
3. Ukubala i-pOH:
\[
\umbhalo{pOH} = 14 – \umbhalo{pH}
\]
\[
\text{pOH} = 14 – 3 = 11
\]
4. Bala ukuhlushwa kwama-ion e-hydroxide:
\[
[\umbhalo{OH}^-] = 10^{-\umbhalo{pOH}}
\]
\[
[\umbhalo{OH}^-] = 10^{-11} \umbhalo{M}
\]
I-Jawaban:
\[
[\text{OH}^-] \text{ kusixazululo se-HF} = 1 \times 10^{-11} \text{M}
\]
Isiphetho
Ukuqonda imiqondo eyisisekelo yama-electrolyte kanye nokuncibilika kubalulekile kumakhemikhali. Ngokwazi ukuthi singabala kanjani izinga le-ionization, i-Ksp, i-pH, kanye ne-pOH, singaxazulula izinkinga ezahlukahlukene ezihlobene nezixazululo ze-electrolyte. Lesi sihloko sixoxe ngezinkinga eziningana zezibonelo kanye nezixazululo zazo ukuze sinikeze ukuqonda okucacile kokuthi zingaxazululwa kanjani izinkinga ezihlobene ne-electrolyte. Ngethemba ukuthi le ncazelo iwusizo kubafundi abafuna ukujulisa ulwazi lwabo ngama-electrolyte kumakhemikhali.
