Imibuzo Eyisibonelo Exoxa Nge-pH Yama-Acids Nezisekelo
Uma sikhuluma ngama-asidi nezisekelo, umqondo owodwa obalulekile okumele siwuqonde yi-pH. I-pH iyisilinganiso se-acidity noma i-alkalinity yesisombululo. Enye ifomula esetshenziswa ukunquma i-pH yile:
\[ \umbhalo{pH} = -\umbhalo [H^+] \]
Kule fomula, \([H^+]\) ukuhlushwa kwama-ion e-hydrogen esixazululweni esilinganiswa nge-molarity (\(\text{mol/L}\)). Ngaphezu kwe-pH, sine-\(\text{pOH}\), esetshenziselwa ukunquma ubunjalo besisombululo:
\[ \umbhalo{pOH} = -\umbhalo [OH^-] \]
Ngemuva kwalokho, ubudlelwano phakathi kwe-pH ne-pOH bulawulwa yile nombolo elandelayo:
\[ \text{pH} + \text{pOH} = 14 \]
Ngezansi sizoxoxa ngemibuzo eminingana eyisibonelo mayelana nendlela yokubala i-pH yezixazululo ze-asidi nesisekelo, kanye nezingxoxo zazo.
Isibonelo Umbuzo 1: Ukubala i-pH yesisombululo se-Acid Esiqinile
Umbuzo:
Bala i-pH yesisombululo se-HCl (i-hydrochloric acid) ngokuhlushwa okungu-0,01 M.
Ingxoxo:
I-HCl iyi-asidi enamandla ezohlukana ngokuphelele emanzini:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
Njengoba i-HCl ihlukanisiwe ngokuphelele, ukuhlushwa kwama-ion e-hydrogen \([H^+]\) esixazululweni kuzofana nokuhlushwa kokuqala kwe-HCl, okungu-0,01 M.
\[ [H^+] = 0,01 \, \umbhalo{M} \]
Okulandelayo, sisebenzisa ifomula ye-pH:
\[ \umbhalo{pH} = -\umbhalo [H^+] \]
\[ \umbhalo{pH} = -\umbhalo (0,01) \]
\[ \text{pH} = -\log (10^{-2}) \]
\[ \umbhalo{pH} = 2 \]
Ngakho-ke, i-pH yesisombululo se-HCl esingu-0,01 M ingu-2.
Isibonelo Umbuzo 2: Ukubala i-pH yesisombululo sesisekelo esiqinile
Umbuzo:
Bala i-pH yesisombululo se-NaOH (i-sodium hydroxide) ngokuhlushwa okungu-0,001 M.
Ingxoxo:
I-NaOH iyisisekelo esiqinile esihlukana ngokuphelele emanzini:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]
Ukuhlushwa kwama-ion e-hydroxide \([OH^-]\) esixazululweni kuzofana nokuhlushwa kokuqala kwe-NaOH, okungukuthi u-0,001 M.
\[ [OH^-] = 0,001 \, \umbhalo{M} \]
Okulandelayo, sibala i-pOH:
\[ \umbhalo{pOH} = -\umbhalo [OH^-] \]
\[ \umbhalo{pOH} = -\umbhalo (0,001) \]
\[ \text{pOH} = -\log (10^{-3}) \]
\[ \umbhalo{pOH} = 3 \]
Ngemva kwalokho, sisebenzisa ubudlelwano phakathi kwe-pH ne-pOH:
\[ \text{pH} + \text{pOH} = 14 \]
\[ \umbhalo{pH} + 3 = 14 \]
\[ \umbhalo{pH} = 11 \]
Ngakho-ke, i-pH yesisombululo se-NaOH esingu-0,001 M ingu-11.
Isibonelo Umbuzo 3: Ukubala i-pH yesisombululo se-Acid esibuthakathaka
Umbuzo:
Bala i-pH yesisombululo se-CH3COOH (i-acetic acid) ngokuhlushwa okungu-0,01 M kanye ne-dissociation constant ye-\(K_a = 1,8 \times 10^{-5}\).
Ingxoxo:
Uma sesinayo i-asidi ebuthakathaka, njenge-acetic acid, engahlukani ngokuphelele, kumele sisebenzise i-acid dissociation constant (\(K_a\)) ukuthola ukuhlushwa kwama-ion e-H+ esixazululweni.
Isilinganiso sokuhlukaniswa kwe-acetic acid emanzini:
\[ \text{CH}_3\text{COOH} \leftrightarrow \text{H}^+ + \text{CH}_3\text{COO}^- \]
I-dissociation constant (\(K_a\)):
\[ K_a = \frac{[H^+] [\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \]
Ake sicabange ukuthi ukuhlushwa kwama-ion e-hydrogen nama-ion e-acetate kungu-\(x\), bese kuthi:
\[ K_a = \frac{x \cdot x}{0,01 – x} \]
Njengoba i-\(K_a\) incane kakhulu, singacabanga ukuthi i-\(0,01 – x \cishe i-0,01\):
\[ 1,8 \izikhathi eziyi-10^{-5} = \frac{x^2}{0,01} \]
\[ x^2 = 1,8 \izikhathi ezingu-10^{-5} \izikhathi ezingu-0,01 \]
\[ x^2 = 1,8 \izikhathi eziyi-10^{-7} \]
\[ x = \sqrt{1,8 \izikhathi eziyi-10^{-7}} \]
\[ x \cishe 1,34 \izikhathi eziyi-10^{-4} \]
Ngakho-ke, ukuhlushwa kwama-ion e-hydrogen \([H^+]\) kungu-\(1,34 \times 10^{-4} \, \text{M}\).
Okulandelayo, sibala i-pH:
\[ \umbhalo{pH} = -\umbhalo [H^+] \]
\[ \text{pH} = -\log (1,34 \times 10^{-4}) \]
\[ \text{pH} \cishe 3,87 \]
Ngakho-ke, i-pH yesisombululo se-acetic acid esingu-0,01 M cishe ingu-3,87.
Isibonelo Umbuzo 4: Ukubala i-pH yesisombululo se-Weak Base
Umbuzo:
Bala i-pH yesisombululo se-NH3 (ammonia) ngokuhlushwa okungu-0,01 M kanye ne-base dissociation constant \(K_b = 1,8 \times 10^{-5}\).
Ingxoxo:
I-NH3 iyisisekelo esibuthakathaka esingahlukani ngokuphelele. Kumelwe sisebenzise i-base dissociation constant (\(K_b\)) ukuthola ukuhlushwa kwama-OH^- ions esixazululweni.
Ukusabela kokuhlukaniswa kwe-ammonia emanzini:
\[ \text{NH}_3 + \text{H}_2\text{O} \leftrightarrow \text{NH}_4^+ + \text{OH}^- \]
I-base dissociation constant (\(K_b\)):
\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \]
Ake sicabange ukuthi ukuhlushwa kwama-ion e-ammonium nama-ion e-hydroxide kungu-\(x\), bese kuthi:
\[ K_b = \frac{x \cdot x}{0,01 – x} \]
Njengoba i-\(K_b\) incane kakhulu, singacabanga ukuthi i-\(0,01 – x \cishe i-0,01\):
\[ 1,8 \izikhathi eziyi-10^{-5} = \frac{x^2}{0,01} \]
\[ x^2 = 1,8 \izikhathi ezingu-10^{-5} \izikhathi ezingu-0,01 \]
\[ x^2 = 1,8 \izikhathi eziyi-10^{-7} \]
\[ x = \sqrt{1,8 \izikhathi eziyi-10^{-7}} \]
\[ x \cishe 1,34 \izikhathi eziyi-10^{-4} \]
Ngakho-ke, ukuhlushwa kwama-ion e-hydroxide \([OH^-]\) kungu-\(1,34 \times 10^{-4} \, \text{M}\).
Okulandelayo, sibala i-pOH:
\[ \umbhalo{pOH} = -\umbhalo [OH^-] \]
\[ \text{pOH} = -\log (1,34 \times 10^{-4}) \]
\[ \text{pOH} \cishe 3,87 \]
Ngemva kwalokho, sisebenzisa ubudlelwano phakathi kwe-pH ne-pOH:
\[ \text{pH} + \text{pOH} = 14 \]
\[ \umbhalo{pH} + 3,87 = 14 \]
\[ \text{pH} \cishe 10,13 \]
Ngakho-ke, i-pH yesisombululo se-ammonia esingu-0,01 M cishe ingu-10,13.
Isiphetho
Ekufundweni kwe-pH, kubalulekile ukuqonda umehluko phakathi kwama-asidi aqinile nabuthakathaka kanye nezisekelo, nokuthi ngayinye ihlukana kanjani nesisombululo. Lokhu kuthinta ngqo indlela esibala ngayo i-pH yesisombululo esithile. Ukubala i-pH kuhilela ukusetshenziswa kwama-logarithms kanye nezimiso zamakhemikhali eziyisisekelo. Ukuqonda le mibono kungasisiza ezindleleni ezahlukahlukene zansuku zonke zokusebenzisa ikhemistri kanye nebhayoloji.
