Ngā Tauira Pātai e Matapaki ana i te Kaha o ngā Waikawa me ngā Kawenga
He ariā nui te kaha o te waikawa me te kawakore i roto i te matū e whakaatu ana i te whai huatanga o te waikawa, te kawakore rānei ki te tuku i te pūrotene (H⁺) me te tango rānei i te pūrotene. I roto i tēnei kōrero, ka tirohia ētahi tauira rapanga hei āwhina i a tātou ki te mārama ki te kaha o te waikawa me te kawakore, me pēhea hoki te tatau i ō rātou pH me te pOH. I mua i te urunga atu ki ngā kōrero rapanga, me mārama tuatahi tātou ki ngā kaupapa matua o te ariā waikawa me te kawakore.
He Kupu Whakataki ki te Kaha o ngā Waikawa me ngā Kawenga
Ka taea te whakatau i te kaha o te waikawa, o te turanga rānei mā te nui o te ionization i roto i te wairewa. Ko te waikawa kaha he mea kua tata te ionization katoa i roto i te wairewa, ko te waikawa ngoikore ia he iti noa te ionization. Pērā anō hoki mō ngā turanga.
Ngā Waikawa Kaha me ngā Waikawa Ngoikore
Ngā waikawa kaha, pērā i:
– Waikawa hauwaihauora (HCl)
– Waikawa whanariki (H₂SO₄)
– Waikawa hauota (HNO₃)
Kua tata te katote katoa i roto i te wairewa. Hei tauira, ka wehea te HCl i roto i te wai hei H⁺ me te Cl⁻ tata te katoa.
Ngā waikawa ngoikore, pērā i:
– Waikawa acetic (CH₃COOH)
– Waikawa hauwaihau (HF)
– Waikawa haitarika (C₆H₈O₇)
he wāhanga noa iho te whaka-ioniti i roto i te wairewa. Ko te tikanga tēnei, i tētahi wā, ka noho tonu ētahi o ngā ngota waikawa i roto i te āhua ngota, ā, kāore e wehea hei katote.
Ngā Pūtake Kaha me ngā Pūtake Ngoikore
Ngā turanga pakari, pērā i:
– Te hauwai konutai (NaOH)
– Te pāhare pākawa waro (KOH)
– Te konupūmā hauwai (Ca(OH)₂)
Kua tata katoa te whaka-ioniti i roto i te wairewa, ā, he nui ngā iona OH⁻ e puta mai ana.
Ngā turanga ngoikore, pērā i:
– Haukini (NH₃)
– Methylamine (CH₃NH₂)
ka whaka-ionitihia ā-wāhanga noa iho i roto i te wairewa, ā, he iti noa iho ngā iona OH⁻ e puta mai ana.
Ngā Pātai Tauira me te Kōrero
Me whakamātau tātou i ētahi tauira rapanga kia pai ake ai te mārama ki tēnei ariā.
Tauira Pātai 1: Te Tātai i te pH o te Waikawa Kaha
Pātai:
Tātaihia te pH o tētahi otinga HCl 0,1 M.
Kōrero:
He waikawa kaha te HCl, nō reira ka tino ionize i roto i te wairewa:
\[ \kuputuhi{HCl} \rightarrow \kuputuhi{H}^+ + \kuputuhi{Cl}^- \]
Nā te mea kua tino ionized a HCl, he ōrite te kukū o [H⁺] ki te kukū o HCl:
\[ [\kuputuhi{H}^+] = 0,1 \, \kuputuhi{M} \]
\[ \kuputuhi{pH} = -\log[\kuputuhi{H}^+] \]
\[ \kuputuhi{pH} = -\log(0,1) \]
\[ \text{pH} = 1 \]
Nō reira, ko te pH o te otinga HCl 0,1 M ko 1.
Tauira Pātai 2: Te Tātai i te pH o te Waikawa Ngoikore
Pātai:
Tātaihia te pH o tētahi otinga CH₃COOH 0,1 M. E mōhiotia ana ko te Ka (pūmau wehewehe waikawa) o te CH₃COOH ko \(1,8 \times 10^{-5}\).
Kōrero:
Mō ngā waikawa ngoikore, me whai whakaaro ki te nui o te ionization. Ko te whārite taurite mō te wetewete o te waikawa acetic ko:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{COO}^- \]
Mā te whakamahi i a Ka:
\[ \kuputuhi{Ka} = \frac{[\kuputuhi{H}^+][\kuputuhi{CH}_3\kuputuhi{COO}^-]}{[\kuputuhi{CH}_3\kuputuhi{COOH}]} \]
Me kī ko te wehenga [H⁺] = [CH₃COO⁻] = x, ā, ko te [CH₃COOH] tuatahi ko te 0,1 M, kātahi:
\[ 1,8 \times 10^{-5} = \frac{x^2}{0,1 – x} \]
Nā te mea he iti a Ka, ka taea te wareware i a x ki te whakatauritea ki te 0,1:
\[ 1,8 \times 10^{-5} \approx \frac{x^2}{0,1} \]
\[ x^2 = 1,8 \whakareatia ki te 10^{-6} \]
\[ x = \sqrt{1,8 \times 10^{-6}} \]
\[ x \tata ki te 1,34 \whakareatia ki te 10^{-3} \]
Kātahi, [H⁺] ≈ \(1,34 \times 10^{-3} \):
\[ \kuputuhi{pH} = -\log[\kuputuhi{H}^+] \]
\[ \text{pH} = -\log(1,34 \times 10^{-3}) \]
\[ \text{pH} \tata ki te 2,87 \]
Nō reira, ko te pH o te otinga 0,1 M CH₃COOH he tata ki te 2,87.
Tauira Pātai 3: Te Tātai i te pOH me te pH o ngā Pūtake Kaha
Pātai:
Tātaihia te pOH me te pH o tētahi otinga NaOH 0,01 M.
Kōrero:
He turanga kaha te NaOH e tino whakakōkītia ana i roto i te wai:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]
Nā te mea kua tino ionized te NaOH, he ōrite te kukū o [OH⁻] ki te kukū o NaOH:
\[ [\kuputuhi{OH}^-] = 0,01 \, \kuputuhi{M} \]
\[ \kuputuhi{pOH} = -\log[\kuputuhi{OH}^-] \]
\[ \kuputuhi{pOH} = -\log(0,01) \]
\[ \text{pOH} = 2 \]
E mōhiotia ana he hononga te pH me te pOH mā te whārite:
\[ \kuputuhi{pH} + \kuputuhi{pOH} = 14 \]
Nō reira,
\[ \kuputuhi{pH} = 14 – \kuputuhi{pOH} \]
\[ \text{pH} = 14 – 2 \]
\[ \text{pH} = 12 \]
Nō reira, ko te pOH o te otinga NaOH 0,01 M he 2, ā, ko te pH he 12.
Tauira Pātai 4: Te Tātai i te pH o tētahi Pūtake Ngoikore
Pātai:
Tātaihia te pH o tētahi otinga NH₃ 0,1 M. E mōhiotia ana ko te Kb (pūmau wehewehe turanga) o te NH₃ ko \(1,76 \times 10^{-5}\).
Kōrero:
Ka tauhohe te NH₃ i roto i te wai hei whakaputa i te NH₄⁺ me te OH⁻:
\[ \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]
Mā te whakamahi i a Kb,
\[ \kuputuhi{Kb} = \frac{[\kuputuhi{NH}_4^+][\kuputuhi{OH}^-]}{[\kuputuhi{NH}_3]} \]
Me kī ko te wehenga [OH⁻] = [NH₄⁺] = x, ā, ko te [NH₃] tuatahi ko 0,1 M, kātahi:
\[ 1,76 \times 10^{-5} = \frac{x^2}{0,1 – x} \]
Hei whakaaro noa, he iti rawa te x ki te whakatauritea ki te 0,1:
\[ 1,76 \times 10^{-5} \approx \frac{x^2}{0,1} \]
\[ x^2 = 1,76 \whakareatia ki te 10^{-6} \]
\[ x = \sqrt{1,76 \times 10^{-6}} \]
\[ x \tata ki te 1,33 \whakareatia ki te 10^{-3} \]
Kātahi, [OH⁻] ≈ \(1,33 \times 10^{-3}\):
\[ \kuputuhi{pOH} = -\log[\kuputuhi{OH}^-] \]
\[ \text{pOH} = -\log(1,33 \times 10^{-3}) \]
\[ \text{pOH} \tata ki te 2,88 \]
Kātahi ka tatauhia te pH mā te:
\[ \kuputuhi{pH} = 14 – \kuputuhi{pOH} \]
\[ \text{pH} = 14 – 2,88 \]
\[ \text{pH} \tata ki te 11,12 \]
Nō reira, ko te pH o tētahi otinga NH₃ 0,1 M he tata ki te 11,12.
Whakamutunga
He mea nui te mārama ki te kaha o ngā waikawa me ngā kawakore ki te matū. Mā te mārama ki te tatau i te pH me te pOH, me te mōhio ki te tautuhi i ngā waikawa me ngā kawakore ngoikore me te kaha, ka pai ake tō whakarite ki te aro atu ki ngā momo raruraru matū e pā ana ki ngā otinga. Ko te tumanako ka whakamāramahia ēnei ariā e ngā tauira kua whakaratohia, ā, ka whakarato i tētahi turanga pakari mō ētahi atu rangahau.
