Ngā Tauira Pātai e Matapaki ana i te pH o ngā Waikawa me ngā Kawenga
Ina kōrero tātou mō ngā waikawa me ngā kawakore, ko tētahi ariā nui me mārama tātou ko te pH. Ko te pH he ine i te waikawa, i te kawakore rānei o tētahi otinga. Ko tētahi tātai e whakamahia ana hei whakatau i te pH ko:
\[ \text{pH} = -\log [H^+] \]
I roto i tēnei tātai, ko te \([H^+]\) te kukū o ngā iona hauwai i roto i te otinga i inehia i roto i te molarity (\(\text{mol/L}\)). Haunga te pH, kei a tātou anō hoki te \(\text{pOH}\), e whakamahia ana hei whakatau i te taketake o tētahi otinga:
\[ \text{pOH} = -\log [OH^-] \]
Kātahi, ko te whanaungatanga i waenga i te pH me te pOH ka whakahaerea e te whārite e whai ake nei:
\[ \kuputuhi{pH} + \kuputuhi{pOH} = 14 \]
Kei raro nei ka matapakihia e mātou ētahi tauira pātai e pā ana ki te tatau i te pH o ngā wairewa waikawa me te kawakore, me ngā matapakinga mō aua pātai.
Tauira Pātai 1: Te Tātai i te pH o tētahi Waikawa Kaha
Pātai:
Tātaihia te pH o tētahi otinga HCl (waikawa hauwaiora) he 0,01 M te kukū.
Kōrero:
He waikawa kaha te HCl ka memeha rawa i roto i te wai:
\[ \kuputuhi{HCl} \rightarrow \kuputuhi{H}^+ + \kuputuhi{Cl}^- \]
Nā te mea kua tino wehea te HCl, ka rite te kukū o ngā katote hauwai \([H^+]\) i roto i te otinga ki te kukū tīmatanga o te HCl, arā, 0,01 M.
\[ [H^+] = 0,01 \, \kuputuhi{M} \]
Muri iho, ka whakamahia e mātou te tātai pH:
\[ \text{pH} = -\log [H^+] \]
\[ \text{pH} = -\log (0,01) \]
\[ \text{pH} = -\log (10^{-2}) \]
\[ \text{pH} = 2 \]
Nō reira, ko te pH o te otinga HCl 0,01 M he 2.
Tauira Pātai 2: Te Tātai i te pH o tētahi Wairewa Pūtake Kaha
Pātai:
Tātaihia te pH o te otinga NaOH (konutai hauwai) me te kukū o te 0,001 M.
Kōrero:
He turanga kaha te NaOH e memeha rawa ana i roto i te wai:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]
Ka rite te kukū o ngā katote hauwai ([OH^-]) i roto i te otinga ki te kukū tīmatanga o te NaOH, arā, 0,001 M.
\[ [OH^-] = 0,001 \, \kuputuhi{M} \]
Muri iho, ka tatauhia e mātou te pOH:
\[ \text{pOH} = -\log [OH^-] \]
\[ \text{pOH} = -\log (0,001) \]
\[ \text{pOH} = -\log (10^{-3}) \]
\[ \text{pOH} = 3 \]
Ā muri iho, ka whakamahia e mātou te whanaungatanga i waenga i te pH me te pOH:
\[ \kuputuhi{pH} + \kuputuhi{pOH} = 14 \]
\[ \text{pH} + 3 = 14 \]
\[ \text{pH} = 11 \]
Nō reira, ko te pH o te otinga NaOH 0,001 M he 11.
Tauira Pātai 3: Te Tātai i te pH o tētahi Waikawa Waikawa Ngoikore
Pātai:
Tātaihia te pH o tētahi otinga CH3COOH (waikawa acetic) me te kukū o te 0,01 M me te pūmau wehenga o \(K_a = 1,8 \times 10^{-5}\).
Kōrero:
Kia whiwhi tātou i tētahi waikawa ngoikore, pērā i te waikawa acetic, e kore e tino wehea, me whakamahi tātou i te pūmau wehenga waikawa (\(K_a\)) hei kimi i te kukū o ngā katote H+ i roto i te otinga.
Ko te whārite mō te wehenga o te waikawa acetic i roto i te wai:
\[ \text{CH}_3\text{COOH} \leftrightarrow \text{H}^+ + \text{CH}_3\text{COO}^- \]
Pūmau wehewehe (\(K_a\)):
\[ K_a = \frac{[H^+] [\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \]
Me kī tātou ko te kukū o ngā iona hauwai me ngā iona ātete ko \(x\), kātahi:
\[ K_a = \frac{x \cdot x}{0,01 – x} \]
Nā te mea he tino iti a \(K_a\), ka taea e tātou te whakaaro ko \(0,01 – x \approx 0,01\):
\[ 1,8 \times 10^{-5} = \frac{x^2}{0,01} \]
\[ x^2 = 1,8 \times 10^{-5} \times 0,01 \]
\[ x^2 = 1,8 \whakareatia ki te 10^{-7} \]
\[ x = \sqrt{1,8 \times 10^{-7}} \]
\[ x \tata ki te 1,34 \whakareatia ki te 10^{-4} \]
Nō reira, ko te kukū o ngā iona hauwai \([H^+]\) ko \(1,34 \times 10^{-4} \, \text{M}\).
Muri iho, ka tatauhia e mātou te pH:
\[ \text{pH} = -\log [H^+] \]
\[ \text{pH} = -\log (1,34 \times 10^{-4}) \]
\[ \text{pH} \tata ki te 3,87 \]
Nō reira, ko te pH o te otinga waikawa acetic M 0,01 he tata ki te 3,87.
Tauira Pātai 4: Te Tātai i te pH o tētahi Otinga Pūtake Ngoikore
Pātai:
Tātaihia te pH o tētahi otinga o te NH3 (haukini) me te kukū o te 0,01 M me te pūmau wehenga turanga \(K_b = 1,8 \times 10^{-5}\).
Kōrero:
He turanga ngoikore te NH3 e kore e tino wehea. Me whakamahi tātou i te pūmau wehenga turanga (\(K_b\)) hei kimi i te kukū o ngā iona OH^- i roto i te wairewa.
Te tauhohenga wehenga o te haukini i roto i te wai:
\[ \text{NH}_3 + \text{H}_2\text{O} \leftrightarrow \text{NH}_4^+ + \text{OH}^- \]
Pūmau wehewehe turanga (\(K_b\)):
\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \]
Me kī tātou ko te kukū o ngā iona ammonia me ngā iona hauwai ko \(x\), kātahi:
\[ K_b = \frac{x \cdot x}{0,01 – x} \]
Nā te mea he tino iti a \(K_b\), ka taea e tātou te whakaaro ko \(0,01 – x \approx 0,01\):
\[ 1,8 \times 10^{-5} = \frac{x^2}{0,01} \]
\[ x^2 = 1,8 \times 10^{-5} \times 0,01 \]
\[ x^2 = 1,8 \whakareatia ki te 10^{-7} \]
\[ x = \sqrt{1,8 \times 10^{-7}} \]
\[ x \tata ki te 1,34 \whakareatia ki te 10^{-4} \]
Nō reira, ko te kukū o ngā katote hauwai \([OH^-]\) ko \(1,34 \times 10^{-4} \, \text{M}\).
Muri iho, ka tatauhia e mātou te pOH:
\[ \text{pOH} = -\log [OH^-] \]
\[ \text{pOH} = -\log (1,34 \times 10^{-4}) \]
\[ \text{pOH} \tata ki te 3,87 \]
Ā muri iho, ka whakamahia e mātou te whanaungatanga i waenga i te pH me te pOH:
\[ \kuputuhi{pH} + \kuputuhi{pOH} = 14 \]
\[ \text{pH} + 3,87 = 14 \]
\[ \text{pH} \tata ki te 10,13 \]
Nō reira, ko te pH o te otinga haukini 0,01 M he tata ki te 10,13.
Whakamutunga
I roto i te ako i te pH, he mea nui kia mārama ki te rerekētanga i waenga i ngā waikawa me ngā turanga kaha me ngā waikawa ngoikore, me te pehea e wehea ai ia mea i roto i te otinga. Ka pā tika tēnei ki te huarahi e tatau ai tātou i te pH o tētahi otinga. Ko te tatau i te pH e uru ana ki te whakamahi i ngā logarithm me ngā mātāpono matū taketake. Mā te mārama ki ēnei ariā ka āwhina i a tātou i roto i ngā momo tono o ia rā o te matū me te koiora.
