Imibuzo Eyisibonelo Exoxa Ngezixazululo Ze-Buffer
Isixazululo se-buffer, esivame ukubizwa ngokuthi i-buffer, siyisisombululo esigcina i-pH yaso ngisho noma kufakwe inani elincane le-asidi noma isisekelo esinamandla. Leli khono lenza izixazululo ze-buffer zibaluleke kakhulu ezisetshenziswayo zamakhemikhali, zebhayoloji, ngisho nasezimbonini. Kulesi sihloko, sizohlola ezinye izibonelo futhi sixoxe ngezixazululo ze-buffer.
Ukuqonda Izixazululo Ze-Buffer
Izixazululo ze-buffer ngokuvamile zakhiwa ingxube ye-asidi ebuthakathaka kanye nesisekelo sayo esihlanganisiwe noma isisekelo esibuthakathaka kanye ne-asidi yayo ehlanganisiwe. Uma i-asidi noma isisekelo sengezwa kulezi zixazululo, ukusabela okubangelwayo kusiza ukugcina i-pH yesisombululo ngaphakathi kobubanzi obuthile. Kulesi simo, i-pH iyisilinganiso se-asidi noma i-alkalinity yesisombululo.
Umsebenzi kanye nokubaluleka kwezixazululo ze-Buffer
Izixazululo ze-buffer zinendima ebalulekile emikhakheni ehlukahlukene:
1. I-Biochemistry: Ukusabela okuningi kwe-enzyme kanye ne-physiological kwenzeka ku-pH ethile, ngakho-ke izixazululo ze-buffer ziyasiza ekugcineni izimo ezifanele.
2. Imboni: Ezimbonini zokudla nezemithi, ukulawulwa kwe-pH okufanele kubalulekile ekhwalithini nasekuphepheni komkhiqizo.
3. Ukuhlolwa Kwamakhemikhali: Elabhorethri, izixazululo ze-buffer zisetshenziselwa ukulawula i-pH ngesikhathi sokusabela kwamakhemikhali.
Indlela Izixazululo Ze-Buffer Ezisebenza Ngayo
Izixazululo ze-buffer zisebenza ngesimiso sokulingana. Ngesixazululo esakhiwe yi-asidi ebuthakathaka \(HA \) kanye nosawoti wayo wesisekelo ohlanganisiwe \(A^- \), ukungezwa kwenani elincane le-asidi noma isisekelo kuzobangela ukusabela kokuguquguquka ngaphandle kokushintsha okukhulu ekugxilweni kwama-ion \(H^+ \) .
Imibuzo Eyisibonelo Nengxoxo
Ukuze sikuqonde kabanzi, ake sihlaziye imibuzo nezingxoxo ezithile ezihlobene nezixazululo ze-buffer.
Umbuzo 1: Ukubala i-pH yesisombululo se-Acid Buffer
Umbuzo:
Isixazululo se-buffer senziwa ngokuxuba i-0,1 mol ye-acetic acid (CH₃COOH) ne-0,1 mol ye-sodium acetate (CH₃COONa) ku-1 litre yesisombululo. I-dissociation constant ye-acetic acid (\( K_a \)) ingu-\( 1,8 \times 10^{-5} \). Bala i-pH yesisombululo.
Ingxoxo:
Lesi sixazululo se-buffer siqukethe i-asidi ebuthakathaka (CH₃COOH) kanye nesisekelo sayo se-conjugate (CH₃COO⁻). Singasebenzisa i-equation ye-Henderson-Hasselbalch ukubala i-pH yesisombululo se-buffer:
\[ \text{pH} = \text{p}K_a + \log \left( \frac{[A^-]}{[HA]} \right) \]
Kuphi:
\[ \umbhalo{p}K_a = -\log (K_a) \]
\[ \text{p}K_a = -\log (1,8 \times 10^{-5}) \]
\[ \text{p}K_a \cishe 4,74 \]
Njengoba ukugxila kwe-\( [A^-] \) kanye ne-\( [HA] \) kufana (0,1 M):
\[ \text{pH} = 4,74 + \log \left( \frac{[0,1]}{[0,1]} \right) \]
\[ \umbhalo{pH} = 4,74 + \ulogi(1) \]
\[ \umbhalo{pH} = 4,74 \]
Ngakho-ke, i-pH yesisombululo se-buffer ingu-4,74.
Umbuzo 2: Ukwengeza i-Acid ku-Buffer Solution
Umbuzo:
Kuyini ushintsho ku-pH uma i-0,01 mol ye-HCl ingezwa ku-1 litre yesisombululo se-buffer esivela kuMbuzo 1?
Ingxoxo:
Ukwengezwa kwe-HCl kuzokwandisa ukuhlushwa kwe-\( H^+ \) esixazululweni. Ukusabela okwenzekayo yilokhu:
\[ \text{H}^+ + \text{CH}_3\text{COO}^- \rightarrow \text{CH}_3\text{COOH} \]
Sidinga ukubala ukuthi zingaki izinhlayiya ze-\( H^+ \) nokuthi zithinta kanjani ukuhlushwa kwe-\( \text{CH}_3\text{COOH} \) kanye ne-\( \text{CH}_3\text{COO}^- \).
Ekuqaleni:
\[ [\text{CH}_3\text{COO}^-] = 0,1 \, \text{mol} \]
\[ [\text{CH}_3\text{COOH}] = 0,1 \, \text{mol} \]
Ngemva kokwengeza u-0,01 mol we-HCl:
\[ [\text{CH}_3\text{COO}^-] = 0,1 – 0,01 = 0,09 \, \text{mol} \]
\[ [\text{CH}_3\text{COOH}] = 0,1 + 0,01 = 0,11 \, \text{mol} \]
Sebenzisa i-equation ye-Henderson-Hasselbalch ukuze ubale i-pH entsha:
\[ \text{pH} = \text{p}K_a + \log \left( \frac{[A^-]}{[HA]} \right) \]
\[ \text{pH} = 4,74 + \log \left( \frac{0,09}{0,11} \right) \]
\[ \umbhalo{pH} = 4,74 + \ulogi(0,818) \]
\[ \umbhalo{pH} = 4,74 + (-0,088) \]
\[ \umbhalo{pH} = 4,65 \]
Ngakho-ke, i-pH yesisombululo se-buffer ngemva kokufaka i-HCl ishintsha kusuka ku-4,74 kuya ku-4,65.
Umbuzo 3: Ukwengeza Isisekelo Esixazululweni Se-Buffer
Umbuzo:
Kuyini ushintsho ku-pH uma i-0,01 mol ye-NaOH ingezwa ku-1 litre yesisombululo se-buffer esivela kuMbuzo 1?
Ingxoxo:
Ukwengezwa kwe-NaOH kuzonciphisa ukuhlushwa kwe-H⁺ esixazululweni yi-OH⁻ evela ku-NaOH. Ukusabela okwenzekayo:
\[ \text{OH}^- + \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \]
Ekuqaleni:
\[ [\text{CH}_3\text{COOH}] = 0,1 \, \text{mol} \]
\[ [\text{CH}_3\text{COO}^-] = 0,1 \, \text{mol} \]
Ngemva kokwengeza u-0,01 mol we-NaOH:
\[ [\text{CH}_3\text{COOH}] = 0,1 – 0,01 = 0,09 \, \text{mol} \]
\[ [\text{CH}_3\text{COO}^-] = 0,1 + 0,01 = 0,11 \, \text{mol} \]
Sebenzisa i-equation ye-Henderson-Hasselbalch ukuze ubale i-pH entsha:
\[ \text{pH} = \text{p}K_a + \log \left( \frac{[A^-]}{[HA]} \right) \]
\[ \text{pH} = 4,74 + \log \left( \frac{0,11}{0,09} \right) \]
\[ \umbhalo{pH} = 4,74 + \ulogi(1,222) \]
\[ \umbhalo{pH} = 4,74 + 0,087 \]
\[ \umbhalo{pH} = 4,83 \]
Ngakho-ke, i-pH yesisombululo se-buffer ngemuva kokwengeza i-NaOH ishintsha kusuka ku-4,74 kuya ku-4,83.
Umbuzo 4: Ukunquma Ukwakheka Kwe-Buffer Nge-pH Ethile
Umbuzo:
Ukuze wenze ilitha eli-1 lesisombululo se-buffer nge-pH engu-5,0 usebenzisa i-acetic acid (CH₃COOH) kanye ne-sodium acetate (CH₃COONa), mangaki ama-moles engxenye ngayinye adingekayo? \( K_a \) ye-acetic acid ingu \( 1,8 \izikhathi ezingu-10^{-5} \).
Ingxoxo:
Sebenzisa i-equation kaHenderson-Hasselbalch:
\[ \text{pH} = \text{p}K_a + \log \left( \frac{[A^-]}{[HA]} \right) \]
Okokuqala, bala i-pKₐ:
\[ \text{p}K_a = -\log (1,8 \times 10^{-5}) \]
\[ \text{p}K_a \cishe 4,74 \]
Bese ufaka inani le-pH olifunayo:
\[ 5,0 = 4,74 + \log \left( \frac{[A^-]}{[HA]} \right) \]
\[ 0,26 = \log \left( \frac{[A^-]}{[HA]} \right) \]
Ukusebenzisa izakhiwo zama-logarithms:
\[ \frac{[A^-]}{[HA]} = 10^{0,26} \]
\[ \frac{[A^-]}{[HA]} \cishe kube ngu-1,82 \]
Lokhu kusho ukuthi nge-molecule ngayinye ye-CH₃COOH, kudingeka ama-moles angu-1,82 e-CH₃COONa. Kulesi silinganiso ngelitha eli-1:
Isibonelo, ake sithi CH₃COOH = x mol, ngakho CH₃COONa = 1,82x mol. Ivolumu iyonke ingu-(x + 1.82x) = 2,82x mol, njengoba senza ilitha eli-1 le-buffer, khona-ke:
\[x + 1.82x = 1 \]
\[ 2.82x = 1 \]
\[ x = \frac{1}{2.82} \cishe kube ngu-0.355 \]
Ngakho-ke, ukwakheka okudingekayo:
– CH₃COOH: 0,355 mol
– CH₃COONA: 1,82 0,355 ≈ 0,646 mol
Ngakho-ke sidinga i-0,355 mol ye-acetic acid kanye ne-0,646 mol ye-sodium acetate ukwenza ilitha eli-1 lesisombululo se-buffer esine-pH engu-5,0.
Lesi sihloko siveza izibonelo ezahlukahlukene zezinkinga ezihlobene nezixazululo ze-buffer kanye nezingxoxo zazo ukuze kunikezwe ukuqonda okujulile kwalomqondo. Sithemba ukuthi lolu lwazi luzosiza ekuqondeni ukuthi izixazululo ze-buffer zisebenza kanjani kanye nezinhlelo zazo ezimweni ezahlukene.
