Mehlala ea Lipotso Tse Buisanang ka pH ea Li-acid le Metheo
Ha re bua ka li-acid le metheo, khopolo e le 'ngoe ea bohlokoa eo re lokelang ho e utloisisa ke pH. pH ke tekanyo ea asiti kapa alkalinity ea tharollo. Foromo e 'ngoe e sebelisoang ho fumana pH ke:
\[ \mongolo{pH} = -\log [H^+] \]
Ka mokhoa ona, \([H^+]\) ke mahloriso a li-ion tsa haedrojene ka har'a tharollo e lekanngoang ka molarity (\(\text{mol/L}\)). Ntle le pH, re boetse re na le \(\text{pOH}\), e sebelisetsoang ho fumana motheo oa tharollo:
\[ \text{pOH} = -\log [OH^-] \]
Ebe, kamano pakeng tsa pH le pOH e laoloa ke equation e latelang:
\[ \mongolo{pH} + \mongolo{pOH} = 14 \]
Ka tlase re tla buisana ka mehlala e 'maloa ea lipotso mabapi le mokhoa oa ho bala pH ea litharollo tsa asiti le tsa motheo, hammoho le lipuisano tsa tsona.
Mohlala oa Potso ea 1: Ho bala pH ea Tharollo ea Asiti e Matla
Potso:
Bala pH ea tharollo ea HCl (hydrochloric acid) ka mahloriso a 0,01 M.
Puisano:
HCl ke asiti e matla e tla arohana ka ho feletseng ka metsing:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
Kaha HCl e arohane ka ho feletseng, mahloriso a li-ion tsa haedrojene \([H^+]\) ka har'a tharollo a tla tšoana le mahloriso a pele a HCl, e leng 0,01 M.
\[ [H^+] = 0,01 \, \mongolo{M} \]
Ka mor'a moo, re sebelisa foromo ea pH:
\[ \mongolo{pH} = -\log [H^+] \]
\[ \mongolo{pH} = -\log (0,01) \]
\[ \text{pH} = -\log (10^{-2}) \]
\[ \mongolo{pH} = 2 \]
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 (sodium hydroxide) ka mahloriso a 0,001 M.
Puisano:
NaOH ke motheo o tiileng o arohanang ka ho feletseng ka metsing:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]
Khakanyo ea li-ion tsa hydroxide \([OH^-]\) ka har'a tharollo e tla tšoana le khakanyo ea pele ea NaOH, e leng 0,001 M.
\[ [OH^-] = 0,001 \, \mongolo{M} \]
Ka mor'a moo, re bala pOH:
\[ \text{pOH} = -\log [OH^-] \]
\[ \mongolo{pOH} = -\log (0,001) \]
\[ \text{pOH} = -\log (10^{-3}) \]
\[ \mongolo{pOH} = 3 \]
Kamora moo, re sebelisa kamano e pakeng tsa pH le pOH:
\[ \mongolo{pH} + \mongolo{pOH} = 14 \]
\[ \mongolo{pH} + 3 = 14 \]
\[ \mongolo{pH} = 11 \]
Kahoo, pH ea tharollo ea NaOH ea 0,001 M ke 11.
Mohlala Potso ea 3: Ho bala pH ea Tharollo ea Asiti e Fokolang
Potso:
Bala pH ea tharollo ea CH3COOH (acetic acid) ka mahloriso a 0,01 M le constant ea karohano ea \(K_a = 1,8 \times 10^{-5}\).
Puisano:
Hang ha re se re e-na le asiti e fokolang, joalo ka asiti ea acetic, e sa arohaneng ka botlalo, re tlameha ho sebelisa kamehla ea karohano ea asiti (\(K_a\)) ho fumana mahloriso a li-ion tsa H+ ka har'a tharollo.
Tekanyo ea ho arohana ha asiti ea acetic ka metsing:
\[ \text{CH}_3\text{COOH} \leftrightarrow \text{H}^+ + \text{CH}_3\text{COO}^- \]
Ho sa fetoheng ha karohano (\(K_a\)):
\[ K_a = \frac{[H^+] [\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \]
A re nke hore mahloriso a li-ion tsa haedrojene le li-ion tsa acetate ke \(x\), ebe:
\[ K_a = \frac{x \cdot x}{0,01 – x} \]
Kaha \(K_a\) e nyane haholo, re ka nahana hore \(0,01 – x \hoo e ka bang 0,01\):
\[ 1,8 \makgetlo a 10^{-5} = \frac{x^2}{0,01} \]
\[ x^2 = 1,8 \makgetlo a 10^{-5} \makgetlo a 0,01 \]
\[ x^2 = 1,8 \makgetlo a 10^{-7} \]
\[ x = \sqrt{1,8 \makgetlo a 10^{-7}} \]
\[ x \hoo e ka bang 1,34 \makgetlo a 10^{-4} \]
Kahoo, mahloriso a li-ion tsa haedrojene \([H^+]\) ke \(1,34 \times 10^{-4} \, \text{M}\).
Ka mor'a moo, re bala pH:
\[ \mongolo{pH} = -\log [H^+] \]
\[ \text{pH} = -\log (1,34 \makgetlo a 10^{-4}) \]
\[ \text{pH} \hoo e ka bang 3,87 \]
Kahoo, pH ea tharollo ea asiti ea acetic ea 0,01 M e ka ba 3,87.
Mohlala Potso ea 4: Ho bala pH ea Tharollo ea Motheo o Fokolang
Potso:
Bala pH ea tharollo ea NH3 (ammonia) ka mahloriso a 0,01 M le botsitso ba karohano ea motheo \(K_b = 1,8 \makhetlo a 10^{-5}\).
Puisano:
NH3 ke motheo o fokolang o sa arohaneng ka ho feletseng. Re tlameha ho sebelisa motheo o sa fetoheng (\(K_b\)) ho fumana mahloriso a li-ion tsa OH^- tharollong.
Karabelo ea karohano ea ammonia ka metsing:
\[ \text{NH}_3 + \text{H}_2\text{O} \leftrightarrow \text{NH}_4^+ + \text{OH}^- \]
Motheo oa karohano o sa fetoheng (\(K_b\)):
\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \]
A re nke hore mahloriso a li-ion tsa ammonium le li-ion tsa hydroxide ke \(x\), ebe:
\[ K_b = \frac{x \cdot x}{0,01 – x} \]
Kaha \(K_b\) e nyane haholo, re ka nahana hore \(0,01 – x \hoo e ka bang 0,01\):
\[ 1,8 \makgetlo a 10^{-5} = \frac{x^2}{0,01} \]
\[ x^2 = 1,8 \makgetlo a 10^{-5} \makgetlo a 0,01 \]
\[ x^2 = 1,8 \makgetlo a 10^{-7} \]
\[ x = \sqrt{1,8 \makgetlo a 10^{-7}} \]
\[ x \hoo e ka bang 1,34 \makgetlo a 10^{-4} \]
Kahoo, mahloriso a li-ion tsa hydroxide \([OH^-]\) ke \(1,34 \times 10^{-4} \, \text{M}\).
Ka mor'a moo, re bala pOH:
\[ \text{pOH} = -\log [OH^-] \]
\[ \text{pOH} = -\log (1,34 \makgetlo a 10^{-4}) \]
\[ \text{pOH} \hoo e ka bang 3,87 \]
Kamora moo, re sebelisa kamano e pakeng tsa pH le pOH:
\[ \mongolo{pH} + \mongolo{pOH} = 14 \]
\[ \mongolo{pH} + 3,87 = 14 \]
\[ \text{pH} \hoo e ka bang 10,13 \]
Kahoo, pH ea tharollo ea ammonia ea 0,01 M e ka ba 10,13.
Qetello
Thutong ea pH, ho bohlokoa ho utloisisa phapang lipakeng tsa li-acid le metheo e matla le e fokolang, le hore na e 'ngoe le e 'ngoe e arohana joang tharollong. Sena se ama ka ho toba tsela eo re balang pH ea tharollo e fanoeng ka eona. Ho bala pH ho kenyelletsa tšebeliso ea li-logarithm le melao-motheo ea mantlha ea lik'hemik'hale. Ho utloisisa likhopolo tsena ho ka re thusa lits'ebetsong tse fapaneng tsa letsatsi le letsatsi tsa k'hemistri le baeloji.
