Zitsanzo za Mafunso Okhudza pH ya Ma Acids ndi Bases
Tikamalankhula za ma acid ndi ma bases, mfundo imodzi yofunika yomwe tiyenera kumvetsetsa ndi pH. pH ndi muyeso wa acidity kapena alkalinity ya yankho. Fomula imodzi yomwe imagwiritsidwa ntchito kudziwa pH ndi iyi:
\[ \text{pH} = -\log [H^+] \]
Mu fomula iyi, \([H^+]\) ndi kuchuluka kwa ma ayoni a haidrojeni mu yankho loyezedwa mu molarity (\(\text{mol/L}\)). Kuwonjezera pa pH, tilinso ndi \(\text{pOH}\), yomwe imagwiritsidwa ntchito kudziwa kufunikira kwa yankho:
\[ \text{pOH} = -\log [OH^-] \]
Kenako, ubale pakati pa pH ndi pOH umayendetsedwa ndi equation yotsatirayi:
\[ \text{pH} + \text{pOH} = 14 \]
Pansipa tikambirana mafunso angapo okhudza momwe tingawerengere pH ya acid ndi base solutions, pamodzi ndi zokambirana zawo.
Chitsanzo Funso 1: Kuwerengera pH ya Strong Acid Solution
Funso:
Werengani pH ya yankho la HCl (hydrochloric acid) ndi kuchuluka kwa 0,01 M.
Kukambirana:
HCl ndi asidi wamphamvu yomwe imasungunuka kwathunthu m'madzi:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
Popeza HCl yalekanitsidwa kwathunthu, kuchuluka kwa ma hydrogen ions \([H^+]\) mu yankho kudzakhala kofanana ndi kuchuluka koyambirira kwa HCl, komwe ndi 0,01 M.
\[ [H^+] = 0,01 \, \malemba{M} \]
Kenako, timagwiritsa ntchito njira ya pH:
\[ \text{pH} = -\log [H^+] \]
\[ \malemba{pH} = -\log (0,01) \]
\[ \text{pH} = -\log (10^{-2}) \]
\[ \malemba{pH} = 2 \]
Kotero, pH ya yankho la 0,01 M HCl ndi 2.
Chitsanzo Funso 2: Kuwerengera pH ya Yankho Lolimba
Funso:
Werengani pH ya yankho la NaOH (sodium hydroxide) ndi kuchuluka kwa 0,001 M.
Kukambirana:
NaOH ndi maziko olimba omwe amalekanitsidwa kwathunthu m'madzi:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]
Kuchuluka kwa ma ion a hydroxide \([OH^-]\) mu yankho kudzakhala kofanana ndi kuchuluka koyambirira kwa NaOH, komwe ndi 0,001 M.
\[ [OH^-] = 0,001 \, \malemba{M} \]
Kenako, timawerengera pOH:
\[ \text{pOH} = -\log [OH^-] \]
\[ \text{pOH} = -\log (0,001) \]
\[ \text{pOH} = -\log (10^{-3}) \]
\[ \malemba{pOH} = 3 \]
Pambuyo pake, timagwiritsa ntchito ubale pakati pa pH ndi pOH:
\[ \text{pH} + \text{pOH} = 14 \]
\[ \text{pH} + 3 = 14 \]
\[ \malemba{pH} = 11 \]
Kotero, pH ya yankho la 0,001 M NaOH ndi 11.
Chitsanzo Funso 3: Kuwerengera pH ya Wofooka wa Acid Solution
Funso:
Werengerani pH ya yankho la CH3COOH (acetic acid) ndi kuchuluka kwa 0,01 M ndi nthawi yosakanikirana ya \(K_a = 1,8 \times 10^{-5}\).
Kukambirana:
Tikakhala ndi asidi wofooka, monga acetic acid, yomwe simalekanitsidwa kwathunthu, tiyenera kugwiritsa ntchito chokhazikika cha dissociation ya acid (\(K_a\)) kuti tipeze kuchuluka kwa ma H+ ions mu yankho.
Chiyerekezo cha kugawanika kwa acetic acid m'madzi:
\[ \text{CH}_3\text{COOH} \leftrightarrow \text{H}^+ + \text{CH}_3\text{COO}^- \]
Chosasinthika cha kugawanika (\(K_a\)):
\[ K_a = \frac{[H^+] [\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \]
Tiyeni tiyerekeze kuti kuchuluka kwa ma ayoni a haidrojeni ndi ma ayoni a acetate ndi \(x\), ndiye:
\[ K_a = \frac{x \cdot x}{0,01 – x} \]
Popeza \(K_a\) ndi yaying'ono kwambiri, tinganene kuti \(0,01 – x \pafupifupi 0,01\):
\[ 1,8 \nthawi 10^{-5} = \frac{x^2}{0,01} \]
\[ x^2 = 1,8 \nthawi 10^{-5} \nthawi 0,01 \]
\[ x^2 = 1,8 \nthawi 10^{-7} \]
\[ x = \sqrt{1,8 \times 10^{-7}} \]
\[ x \pafupifupi 1,34 \nthawi 10^{-4} \]
Kotero, kuchuluka kwa ma ayoni a haidrojeni \([H^+]\) ndi \(1,34 \times 10^{-4} \, \text{M}\).
Kenako, timawerengera pH:
\[ \text{pH} = -\log [H^+] \]
\[ \text{pH} = -\log (1,34 \times 10^{-4}) \]
\[ \text{pH} \pafupifupi 3,87 \]
Kotero, pH ya yankho la 0,01 M acetic acid ndi pafupifupi 3,87.
Chitsanzo Funso 4: Kuwerengera pH ya Yankho Lofooka
Funso:
Werengerani pH ya yankho la NH3 (ammonia) ndi kuchuluka kwa 0,01 M ndi chokhazikika cha dissociation ya base \(K_b = 1,8 \times 10^{-5}\).
Kukambirana:
NH3 ndi maziko ofooka omwe sangalekanitsidwe kwathunthu. Tiyenera kugwiritsa ntchito chokhazikika cha dissociation base (\(K_b\)) kuti tipeze kuchuluka kwa ma OH^- ions mu yankho.
Kugawanika kwa ammonia m'madzi:
\[ \text{NH}_3 + \text{H}_2\text{O} \leftrightarrow \text{NH}_4^+ + \text{OH}^- \]
Chokhazikika cha kugawanika kwa maziko (\(K_b\)):
\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \]
Tiyeni tiyerekeze kuti kuchuluka kwa ma ayoni a ammonium ndi ma ayoni a hydroxide ndi \(x\), ndiye:
\[ K_b = \frac{x \cdot x}{0,01 – x} \]
Popeza \(K_b\) ndi yaying'ono kwambiri, tingaganize kuti \(0,01 – x \pafupifupi 0,01\):
\[ 1,8 \nthawi 10^{-5} = \frac{x^2}{0,01} \]
\[ x^2 = 1,8 \nthawi 10^{-5} \nthawi 0,01 \]
\[ x^2 = 1,8 \nthawi 10^{-7} \]
\[ x = \sqrt{1,8 \times 10^{-7}} \]
\[ x \pafupifupi 1,34 \nthawi 10^{-4} \]
Kotero, kuchuluka kwa ma hydroxide ions \([OH^-]\) ndi \(1,34 \times 10^{-4} \, \text{M}\).
Kenako, timawerengera pOH:
\[ \text{pOH} = -\log [OH^-] \]
\[ \text{pOH} = -\log (1,34 \times 10^{-4}) \]
\[ \text{pOH} \pafupifupi 3,87 \]
Pambuyo pake, timagwiritsa ntchito ubale pakati pa pH ndi pOH:
\[ \text{pH} + \text{pOH} = 14 \]
\[ \text{pH} + 3,87 = 14 \]
\[ \text{pH} \pafupifupi 10,13 \]
Choncho, pH ya yankho la ammonia la 0,01 M ndi pafupifupi 10,13.
Mapeto
Mu kafukufuku wa pH, ndikofunikira kumvetsetsa kusiyana pakati pa asidi amphamvu ndi ofooka ndi maziko, komanso momwe chilichonse chimagawikira mu yankho. Izi zimakhudza mwachindunji momwe timawerengera pH ya yankho loperekedwa. Kuwerengera pH kumaphatikizapo kugwiritsa ntchito ma logarithms ndi mfundo zoyambira zamankhwala. Kumvetsetsa mfundo izi kungatithandize pakugwiritsa ntchito mankhwala ndi zinthu zamoyo tsiku ndi tsiku.
