Laʻana o nā nīnau kūkākūkā no ka Titration Acid-Base
ʻO ka titration acid-base kekahi o nā ʻano loiloi quantitative maʻamau i ka kemika. Pili kēia ʻenehana i ka hopena ma waena o kahi waikawa a me kahi kumu e hoʻoholo ai i ka nui o kahi hopena. Ma kēia ʻatikala, e kūkākūkā mākou i kekahi mau laʻana o nā pilikia titration acid-base a me kā lākou mau hoʻonā, me ka manaʻo e kōkua iā ʻoe e hoʻomaopopo i nā manaʻo kumu a pehea e hana ai iā lākou.
Nā Manaʻo Kumu o ka Titration Acid-Base
Pili nā titrations acid-base i kahi hopena neutralization ma waena o nā ion hydrogen (H⁺) mai kahi waikawa a me nā ion hydroxide (OH⁻) mai kahi kumu e hana i ka wai (H₂O). ʻO ke kiko like ka kiko kahi i like ai ka helu o nā mole o ka waikawa me ka helu o nā mole o ke kumu i hoʻohui ʻia. I kēia manawa, ua hana piha ʻia ka hopena ma waena o ka waikawa a me ke kumu.
Formula Laulā
ʻO ke ʻano kumu i hoʻohana ʻia i ka titration acid-base:
\[ n_{a} \cdot M_{a} \cdot V_{a} = n_{b} \cdot M_{b} \cdot V_{b} \]
Ma hea:
– ʻo \( n_{a} \) ka valence o ka waikawa,
– ʻO \( M_{a} \) ka molarity o ka waikawa,
– ʻO \( V_{a} \) ka nui o ka waikawa,
– ʻO \( n_{b} \) ke kumu valence,
– ʻO \( M_{b} \) ka molarity o ke kumu,
– ʻO \( V_{b} \) ka nui o ke kumu.
Pepa Litmus a me nā Hōʻailona
I ka hana titration, hoʻohana pinepine ʻia nā ʻōkuhi acid-base e like me ka phenolphthalein a i ʻole methyl orange e hoʻoholo ai i ke kiko like. Hoʻololi kēia mau ʻōkuhi i ke kala ma kahi pH kikoʻī, e hōʻike ana ua hiki i ke kiko like.
Nā Nīnau Laʻana a me ke Kūkākūkā
Laʻana Nīnau 1: Titration o ka Waikawa Ikaika a me ke Kumu Ikaika
Nīnau:
Ua ʻike ʻia ua hoʻohuihui ʻia he 25 mL o ka hopena 0,1 M HCl me ka hopena 0,1 M NaOH. He aha ka nui o ka NaOH e pono ai e hiki i ke kiko like?
Kūkākūkā:
ʻO ke ʻano o ka hopena ma waena o HCl a me NaOH penei:
\[ HCl + NaOH \rightarrow NaCl + H_2O \]
KaʻAnuʻu Hana 1: E hoʻoholo i nā mole o H⁺ mai HCl
\[ n_{\text{HCl}} = M_{\text{HCl}} \times V_{\text{HCl}} \]
\[ n_{\text{HCl}} = 0,1 \times 0,025 = 0,0025 \text{ mol} \]
KaʻAnuʻu Hana 2: E hoʻoholo i ka nui o ka NaOH e pono ai
No ka mea, he 1:1 ka lakio o HCl a me NaOH i ka hopena (n = 1):
\[ n_{\text{HCl}} = n_{\text{NaOH}} \]
\[ M_{\text{NaOH}} \times V_{\text{NaOH}} = 0,0025 \text{ mol} \]
\[ V_{\text{NaOH}} = \frac{0,0025 \text{ mol}}{0,1 \text{ M}} = 0,025 \text{ L} = 25 \text{ mL} \]
No laila, ʻo ka nui o ka NaOH e pono ai he 25 mL.
Laʻana Nīnau 2: Titration o ka Waikawa Nāwaliwali a me ke Kumu Ikaika
Nīnau:
Ua hoʻohuihui ʻia he 50 mL o ka 0,1 M acetic acid (CH₃COOH) me ka 0,1 M NaOH solution. He aha ka nui o NaOH e pono ai e hiki i ke kiko like?
Kūkākūkā:
ʻO ke ʻano o ka hopena ma waena o ka waikawa acetic a me ka NaOH penei:
\[ CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O \]
KaʻAnuʻu Hana 1: E hoʻoholo i nā mole o H⁺ mai CH₃COOH
\[ n_{\text{CH}_3\text{COOH}} = M_{\text{CH}_3\text{COOH}} \times V_{\text{CH}_3\text{COOH}} \]
\[ n_{\text{CH}_3\text{COOH}} = 0,1 \times 0,05 = 0,005 \text{ mol} \]
KaʻAnuʻu Hana 2: E hoʻoholo i ka nui o ka NaOH e pono ai
No ka mea, ʻo ka lakio mole ma waena o CH₃COOH a me NaOH he 1:1:
\[ n_{\text{CH}_3\text{COOH}} = n_{\text{NaOH}} \]
\[ M_{\text{NaOH}} \times V_{\text{NaOH}} = 0,005 \text{ mol} \]
\[ V_{\text{NaOH}} = \frac{0,005 \text{ mol}}{0,1 \text{ M}} = 0,05 \text{ L} = 50 \text{ mL} \]
No laila, ʻo ka nui o ka NaOH e pono ai he 50 mL.
Laʻana Nīnau 3: Titration Polyprotic o nā waikawa ikaika a me nā kumu
Nīnau:
Ua ʻike ʻia ua hoʻohuihui ʻia he 40 mL o ka hopena 0,05 M H₂SO₄ me ka hopena 0,1 M KOH. He aha ka nui o KOH e pono ai e hiki i nā kiko like mua a me ka lua?
Kūkākūkā:
He waikawa diprotic ka H₂SO₄ e hiki ke hoʻokuʻu i ʻelua mau ion H⁺. ʻElua pae ka hopena me KOH:
\[ H_2SO_4 + 2KOH \rightarrow K_2SO_4 + 2H_2O \]
KaʻAnuʻu Hana 1: E hoʻoholo i nā mole o H⁺ mai H₂SO₄
\[ n_{\text{H}_2\text{SO}_4} = M_{\text{H}_2\text{SO}_4} \times V_{\text{H}_2\text{SO}_4} \]
\[ n_{\text{H}_2\text{SO}_4} = 0,05 \times 0,04 = 0,002 \text{ mol} \]
KaʻAnuʻu Hana 2: E hoʻoholo i ka nui o KOH e pono ai no kēlā me kēia pae
I ka titration mua (e hoʻokuʻu ana i hoʻokahi H⁺):
\[ n_{\text{H}_2\text{SO}_4} = n_{\text{KOH}} \]
\[ M_{\text{KOH}} \times V_{\text{KOH}} = 0,002 \text{ mol} \]
\[ V_{\text{KOH}}_{\text{first}} = \frac{0,002 \text{ mol}}{0,1 \text{ M}} = 0,02 \text{ L} = 20 \text{ mL} \]
I ka lua o ka titration (e wehe ana i ʻelua H⁺) ʻelua manawa e pono ai nā mole ma mua:
\[ n_{\text{KOH}}_{\text{kekona}} = 2 \times 0,002 \text{ mol} = 0.004 \text{ mol} \]
\[ V_{\text{KOH}}_{\text{kekona}} = \frac{0.004 \text{ mol}}{0,1 \text{ M}} = 0,04 \text{ L} = 40 \text{ mL} \]
No laila, ʻo ka nui o KOH e pono ai no ke kiko like mua he 20 mL, a no ke kiko like ʻelua he 40 mL.
Ka hopena
Ma o nā hiʻohiʻona titration acid-base he nui ma luna, ua aʻo mākou pehea e hoʻoholo ai i ka nui o ka hopena e pono ai e hiki i ke kiko like i nā ʻano titration like ʻole. He mea nui e hoʻolohe mau i ka stoichiometry reaction a hoʻohana pololei i ke ʻano titration kumu e loaʻa ai nā hopena pololei. ʻO ka hoʻomaʻamaʻa mau ʻana me nā ʻano pilikia like ʻole e kōkua i ka hoʻoikaika ʻana i kou ʻike i nā titration acid-base i ka kemika analytical.
