Mehlala ea Lipotso tse Buisanang ka Thepa le Mehopolo ea Li-Acid le Metheo
Thepa le mehopolo ea li-acid le li-base ke lihlooho tsa bohlokoa k'hemistri, ka ts'ebeliso e pharaletseng masimong a fapaneng, ho kenyeletsoa indasteri, khemisi le biochemistry. Ho utloisisa mehopolo ea mantlha le ts'ebeliso ea thepa ea acid-base ho bohlokoa bakeng sa baithuti le bafuputsi. Sengoloa sena se tla tšohla mehlala e 'maloa ea mathata mabapi le thepa le mehopolo ea li-acid le li-base, hammoho le litharollo tsa tsona, ho fa babali setšoantšo se hlakileng haholoanyane.
Mehopolo ea Motheo ea Li-acid le Metheo
Pele re hlahloba lipotso le lipuisano tsa tsona, ha re hlahlobeng likhopolo tse ling tsa motheo mabapi le li-acid le li-base.
Bolila:
– Li-acid ke metsoako e ka fanang ka li-ion tsa haedrojene (H⁺) ka har'a tharollo.
– Li-asidi li na le thepa ea asiti, mohlala, tatso e bolila (joalo ka asene) le bokhoni ba ho arabela le litšepe.
Bass:
– Motheo ke motsoako o ka amohelang li-ion tsa haedrojene kapa oa tsamaisa li-ion tsa hydroxide (OH⁻) ka har'a tharollo.
– Metheo e na le thepa ea motheo, joalo ka tatso e babang le sebopeho se thellang (joalo ka sesepa).
Khopolo-taba ea Motheo oa Asiti:
– Khopolo-taba ea Arrhenius: Li-asidi li hlahisa H⁺ 'me metheo e hlahisa OH⁻ ka tharollo ea metsi.
– Khopolo-taba ea Brønsted-Lowry: Li-asidi ke bafani ba H⁺ 'me metheo ke baamoheli ba H⁺.
– Khopolo-taba ea Lewis: Li-acid ke li-elektrone para tse amohelang 'me metheo ke bafani ba li-elektrone para.
Lipotso le Lipuisano tsa Mehlala
Mohlala oa Potso ea 1: Ho bala pH ea Tharollo ea Asiti e Matla
Potso:
Bala pH ea tharollo ea 0,01 M HCl.
Puisano:
HCl ke asiti e matla e tla arohana ka ho feletseng tharollong ho hlahisa H⁺ le Cl⁻. Khatello ea H⁺ e tšoana le khatello ea pele ea HCl.
\[ [H⁺] = 0,01 \, M \]
pH e hlalosoa e le logarithm e mpe ea motheo oa 10 ea mahloriso a ion ea hydrogen:
\[ \text{pH} = -\log[H⁺] \]
\[ \mongolo{pH} = -\log(0,01) \]
\[ \mongolo{pH} = 2 \]
Kahoo, pH ea tharollo ea 0,01 M HCl ke 2.
Mohlala oa Potso ea 2: Ho bala pH ea Tharollo ea Motheo o Matla
Potso:
Bala pH ea tharollo ea NaOH ea 0,001 M.
Puisano:
NaOH ke motheo o tiileng o tla arohana ka ho feletseng tharollong, o hlahise Na⁺ le OH⁻. Khatello ea OH⁻ e tšoana le khatello ea pele ea NaOH.
\[ [OH⁻] = 0,001 \, M \]
PoH e hlalosoa e le logarithm e mpe ea motheo oa 10 ea mahloriso a ion ea hydroxide:
\[ \mongolo{pOH} = -\log[OH⁻] \]
\[ \mongolo{pOH} = -\log(0,001) \]
\[ \mongolo{pOH} = 3 \]
pH le pOH li amana ka equation:
\[ \mongolo{pH} + \mongolo{pOH} = 14 \]
\[ \mongolo{pH} = 14 – 3 \]
\[ \mongolo{pH} = 11 \]
Kahoo, pH ea tharollo ea NaOH ea 0,001 M ke 11.
Mohlala oa 3: Ho bala pH ea Asiti e Fokolang
Potso:
Bala pH ea tharollo ea 0,1 M acetic acid (CH₃COOH), e nang le constant ea karohano ea asiti (Ka) ea \( 1,8 \makhetlo a 10^{-5} \).
Puisano:
Li-acid tse fokolang ha li arohane ka botlalo tharollong. Ho bala pH, re tlameha ho qala ka ho fumana mahloriso a li-ion tsa H⁺ re sebelisa boleng ba Ka.
\[ CH_3COOH \rightleftharpoons H^+ + CH_3COO^- \]
Haeba \( x \) e le mahloriso a H⁺ a hlahisitsoeng:
\[ Ka = \frac{{[H^+][CH_3COO^-]}}{{[CH_3COOH]}} \]
\[ Ka = \frac{{x \cdot x}}{{0,1 – x}} \]
Kaha ke asiti e fokolang, nahana hore \( x \) e nyane haholo hoo \( 0,1 – x \hoo e ka bang 0,1 \):
\[ 1,8 \makgetlo a 10^{-5} = \frac{x^2}{0,1} \]
\[ x^2 = 1,8 \makgetlo a 10^{-6} \]
\[ x = \sqrt{1,8 \makgetlo a 10^{-6}} \]
\[ x \hoo e ka bang 1,34 \makgetlo a 10^{-3} \]
Khatello ea li-ion tsa H⁺ e ka ba \( 1,34 \makhetlo a 10^{-3} \):
\[ \mongolo{pH} = -\log[H^+] \]
\[ \text{pH} = -\log(1,34 \makgetlo a 10^{-3}) \]
\[ \text{pH} \hoo e ka bang 2,87 \]
Kahoo, pH ea tharollo ea 0,1 M CH₃COOH e ka ba 2,87.
Mohlala oa Potso ea 4: Ho bala pH ea Motheo o Fokolang
Potso:
Bala pH ea tharollo ea ammonia ea 0,05 M (NH₃), e nang le botsitso ba karohano ea motheo (Kb) ea \( 1,8 \makhetlo a 10^{-5} \).
Puisano:
Metheo e fokolang ha e arohane ka botlalo tharollong. Ho bala pH, re tlameha ho qala ka ho fumana mahloriso a li-ion tsa OH⁻ re sebelisa boleng ba Kb.
\[ NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- \]
Haeba \( x \) e le mahloriso a OH⁻ a hlahisitsoeng:
\[ Kb = \frac{{[NH_4^+][OH^-]}}{{[NH_3]}} \]
\[ Kb = \frac{{x \cdot x}}{{0,05 – x}} \]
Nka hore \( x \) e nyane haholo hoo \( 0,05 – x \hoo e ka bang 0,05 \):
\[ 1,8 \makgetlo a 10^{-5} = \frac{x^2}{0,05} \]
\[ x^2 = 9,0 \makgetlo a 10^{-7} \]
\[ x = \sqrt{9,0 \makgetlo a 10^{-7}} \]
\[ x \hoo e ka bang 9,49 \makgetlo a 10^{-4} \]
Khatello ea li-ion tsa OH⁻ e ka ba \( 9,49 \makhetlo a 10^{-4} \):
\[ \mongolo{pOH} = -\log[OH^-] \]
\[ \text{pOH} = -\log(9,49 \makgetlo a 10^{-4}) \]
\[ \text{pOH} \hoo e ka bang 3,02 \]
pH le pOH li amana ka equation:
\[ \mongolo{pH} + \mongolo{pOH} = 14 \]
\[ \mongolo{pH} = 14 – 3,02 \]
\[ \text{pH} \hoo e ka bang 10,98 \]
Mohlala oa Potso ea 5: Ho se nke lehlakore ha li-acid le metheo
Potso:
Ke bolumo bofe ba tharollo ea NaOH ea 0,1 M bo hlokahalang ho fokotsa 50 mL ea tharollo ea HCl ea 0,1 M?
Puisano:
Karabelo ea ho se nke lehlakore pakeng tsa asiti le motheo e hlahisa metsi le letsoai:
\[ HCl + NaOH \ motsu NaCl + H_2O \]
Palo ea li-moles tsa HCl:
\[ n_{HCl} = Molarity \mehla Bolumo \]
\[ n_{HCl} = 0,1 \, M \makgetlo 0,05 \, L \]
\[ n_{HCl} = 0,005 \, mol \]
Hobane mole e 'ngoe le e 'ngoe ea HCl e arabela le mole e le 'ngoe ea NaOH, palo ea li-moles tsa NaOH e hlokahalang e ea tšoana, e leng li-moles tse 0,005.
Bophahamo ba NaOH bo hlokahalang:
\[ n_{NaOH} = Molarity_{NaOH} \methapo Bolumo \]
\[ Moqolo_{NaOH} = \frac{n_{NaOH}}{Molarity_{NaOH}} \]
\[ Moqolo_{NaOH} = \frac{0,005 \, mol}{0,1 \, M} \]
\[ Moqolo_{NaOH} = 0,05 \, L \]
\[ Bolumo_{NaOH} = 50 \, mL \]
Kahoo, ho hlokahala 50 mL ea tharollo ea NaOH ea 0,1 M ho fokotsa 50 mL ea tharollo ea HCl ea 0,1 M.
Ka ho buisana ka lipotso tse ka holimo, re tšepa hore kutloisiso ea rona ea litšobotsi le mehopolo ea li-acid le metheo e hlakile haholoanyane. Litšobotsi le mehopolo ena li bohlokoa haholo lits'ebetsong tse fapaneng tsa k'hemistri tsa letsatsi le letsatsi, ho tloha tlhahlobong ea laboratori ho ea indastering.
