Tambayoyi da Tattaunawa game da Arrhenius Acids da Bases
Acids da tushe muhimman ra'ayoyi ne a fannin ilmin sunadarai. Ɗaya daga cikin shahararrun ka'idoji da ke bayyana halayen acid da tushe shine ka'idar Arrhenius. A cewar Arrhenius, acid wani mahaɗi ne wanda, idan ya narke a cikin ruwa, yana samar da ions na H⁺ (ko H₃O⁺), yayin da tushe wani mahaɗi ne wanda, idan ya narke a cikin ruwa, yana samar da ions na OH⁻. A cikin wannan labarin, za mu tattauna misalai da yawa na matsaloli game da acid da tushe bisa ga ka'idar Arrhenius, tare da mafita.
Misali Tambaya ta 1
Tambaya:
Ana narkar da ruwan sinadarin hydrochloric acid (HCl) mai nauyin 0,1 M a cikin ruwa. A ƙididdige yawan sinadarin H⁺ a cikin ruwan.
Tattaunawa:
A cewar ka'idar Arrhenius, HCl sinadari ne saboda yana samar da ions na H⁺ idan aka narkar da su a cikin ruwa. Daidaiton ionization na HCl a cikin ruwa shine kamar haka:
\[ \text{HCl} \zuwa \text{H}^+ + \text{Cl}^- \]
Yawan HCl na farko shine 0,1 M. Tunda HCl wani sinadari ne mai ƙarfi wanda ke yin ion gaba ɗaya a cikin ruwan da ke narkewa, kowace kwayar halittar HCl za ta samar da ion H⁺ ɗaya da ion Cl⁻ ɗaya. Saboda haka, yawan ions H⁺ a cikin ruwan shi ma yayi daidai da yawan HCl, wato:
\[ [\text{H}^+] = 0,1 \text{ M} \]
Don haka, yawan ions na H⁺ a cikin maganin HCl na 0,1 M shine 0,1 M.
Misali Tambaya ta 2
Tambaya:
Lissafa pH na maganin nitric acid (HNO₃) na 0,01 M.
Tattaunawa:
HNO₃ wani sinadari ne mai ƙarfi wanda ke yin ion gaba ɗaya a cikin ruwa, don haka lissafin ionization na HNO₃ shine kamar haka:
\[ \text{HNO}_3 \zuwa \text{H}^+ + \text{NO}_3^- \]
Yawan HNO₃ da aka bayar shine 0,01 M. Tunda HNO₃ acid ne mai ƙarfi, kowace kwayar HNO₃ za ta samar da ion H⁺ guda ɗaya. Saboda haka, yawan ions H⁺ a cikin maganin daidai yake da yawan HNO₃, wato:
\[ [\text{H}^+] = 0,01 \text{ M} \]
Ana iya ƙididdige pH ta amfani da dabarar:
\[ \text{pH} = -\log [\text{H}^+] \]
Ta hanyar haɗa yawan ions na H⁺ cikin dabarar, muna samun:
\[ \text{pH} = -\log (0,01) = -\log (10^{-2}) = 2 \]
Don haka, pH na maganin 0,01 M HNO₃ shine 2.
Misali Tambaya ta 3
Tambaya:
Kayyade yawan ions na OH⁻ a cikin maganin NaOH na 0,05 M.
Tattaunawa:
NaOH tushe ne mai ƙarfi wanda ke yin ion gaba ɗaya a cikin ruwa, don haka lissafin ionization na NaOH kamar haka:
\[ \text{NaOH} \to \text{Na}^+ + \rubutu{OH}^- \]
Yawan NaOH da aka bayar shine 0,05 M. Tunda NaOH tushe ne mai ƙarfi, kowace ƙwayar NaOH za ta samar da ion OH⁻ guda ɗaya. Saboda haka, yawan ions OH⁻ a cikin maganin daidai yake da yawan NaOH, wato:
\[ [\text{OH}^-] = 0,05 \text{ M} \]
Don haka, yawan ions OH⁻ a cikin maganin NaOH na M 0,05 shine 0,05 M.
Misali Tambaya ta 4
Tambaya:
Lissafa pH na maganin 0,01 M KOH.
Tattaunawa:
KOH tushe ne mai ƙarfi wanda ke yin ion gaba ɗaya a cikin ruwa. Daidaiton ionization na KOH a cikin ruwa shine kamar haka:
\[ \text{KOH} \zuwa \text{K}^+ + \text{OH}^- \]
Yawan KOH da aka bayar shine 0,01 M. Tunda KOH tushe ne mai ƙarfi, kowace ƙwayar KOH za ta samar da ion OH⁻ guda ɗaya. Saboda haka, yawan ions OH⁻ a cikin maganin daidai yake da yawan KOH, wato:
\[ [\text{OH}^-] = 0,01 \text{ M} \]
pH ma'auni ne na yawan ions na H⁺ a cikin wani bayani. Za mu iya amfani da alaƙar da ke tsakanin pH, ions na H⁺, da ions na OH⁻ da samfurin ions na ruwa (Kw) ya bayar:
\[ \text{Kw} = [\text{H}^+][\text{OH}^-] = 1 \times 10^{-14} \]
Da sanin darajar [OH⁻], za mu iya ƙididdige [H⁺]:
\[ [\text{H}^+] = \frac{\text{Kw}}{[\text{OH}^-]} = \frac{1 \times 10^{-14}}{0,01} = 1 \times 10^{-12} \text{ M} \]
Ana iya ƙididdige pH ta amfani da dabarar:
\[ \text{pH} = -\log [\text{H}^+] \]
Ta hanyar haɗa yawan ions na H⁺ cikin dabarar, muna samun:
\[ \text{pH} = -\log (sau 1 10^{-12}) = 12 \]
Don haka, pH na maganin 0,01 M KOH shine 12.
Misali Tambaya ta 5
Tambaya:
Maganin yana da pH = 3. Lissafa yawan ions na H⁺ a cikin maganin.
Tattaunawa:
pH ma'auni ne na yawan ions na H⁺ a cikin wani bayani kuma ana iya bayyana shi ta hanyar dabarar:
\[ \text{pH} = -\log [\text{H}^+] \]
Don ƙididdige yawan ions na H⁺, muna canza dabarar zuwa:
\[ [\text{H}^+] = 10^{-\text{pH}} \]
Ta hanyar haɗa ƙimar pH da aka bayar, mun sami:
\[ [\text{H}^+] = 10^{-3} = 0,001 \text{ M} \]
Don haka, yawan ions na H⁺ a cikin maganin da ke da pH = 3 shine 0,001 M.
Misali Tambaya ta 6
Tambaya:
Maganin NH₄OH yana da yawan sinadarin 0,1 M. A ƙididdige pH na maganin idan ƙimar Kb ga NH₄OH ita ce 1,8 x 10⁻⁵.
Tattaunawa:
NH₄OH tushe ne mai rauni wanda baya yin ion gaba ɗaya a cikin ruwa. Daidaiton ionization na NH₄OH a cikin ruwa shine kamar haka:
\[ \text{NH}_4\text{OH} \leftrightarrow \text{NH}_4^+ + \text{OH}^- \]
Ana kiran ma'aunin daidaito na ionization na NH₄OH Kb kuma ana bayar da shi ta hanyar:
\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_4\text{OH}} \]
Tunda NH₄OH tushe ne mai rauni, muna amfani da Kb don ƙididdige yawan ions na OH⁻. Idan muka ɗauka cewa \(x\) shine yawan NH₄OH mai ionized, lissafin ionization zai zama:
\[ K_b = \frac{x^2}{0,1 – x} \approx \frac{x^2}{0,1} \]
Tunda ƙimar Kb ƙanƙanta ce sosai, muna ɗauka cewa \(0,1 – x \approx 0,1\):
\[ sau 1,8 10^{-5} = \frac{x^2}{0,1} \]
\[ x^2 = 1,8 \sau 10^{-6} \]
\[ x = \sqrt{1,8 \sau 10^{-6}} \]
\[ x \kimanin 1,34 \sau 10^{-3} \text{ M} \]
Don haka, yawan ions na OH⁻ shine \(1,34 \times 10^{-3} \text{ M}\).
Daga wannan yawan, za mu iya ƙididdige yawan ions na H⁺ ta amfani da Kw:
\[ [\text{H}^+] = \frac{1 \times 10^{-14}}{1,34 \times 10^{-3}} \]
\[ [\text{H}^+] = 7,46 \sau 10^{-12} \text{ M} \]
pH:
\[ \text{pH} = -\log (7,46 \sau 10^{-12}) \]
\[ \text{pH} \approx 11,13 \]
Don haka, pH na maganin 0,1 M NH₄OH kusan 11,13 ne.
Kammalawa
Ka'idar Arrhenius tana ba da tushe mai mahimmanci don fahimtar yadda acid da tushe ke aiki a cikin mafita. A cikin wannan tattaunawar, mun duba misalai da yawa na matsaloli waɗanda ke nuna yadda za a iya amfani da waɗannan ra'ayoyi don ƙididdige yawan ions na H⁺ da OH⁻ da kuma pH na mafita daban-daban. Fahimtar wannan ka'idar yana da mahimmanci ga duk wanda ke neman aiki a fannin sinadarai, ko a matakin sakandare ko na ilimi mai zurfi.
