14.6: Polyprotic Acids

Malengo ya kujifunza

Mwishoni mwa sehemu hii, utaweza:

• Kupanua dhana awali ilianzisha msawazo kwa asidi na besi ambayo inaweza kuchangia au kukubali proton zaidi ya moja

Acids huwekwa na idadi ya protoni kwa molekuli ambayo wanaweza kuacha katika mmenyuko. Asidi kama vile HCl, HNO ₃, na HCN ambazo zina atomi moja ya hidrojeni ionizable katika kila molekuli huitwa asidi monoprotiki. Athari zao na maji ni:

$$\begin{array}{c}\text{HCl}(aq)+{\text{H}}_2\text{O}(l)⟶{\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{Cl}}^{\text{-}}(aq)\\ {\text{HAPANA}}_3(aq)+{\text{H}}_2\text{O}(l)⟶{\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{HAPANA}}_3{}^{\text{-}}(aq)\\ \text{HCN}(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{CN}}^{\text{-}}(aq)\end{array}$$

Japokuwa ina atomi nne za hidrojeni, asidi asetiki, CH ₃ CO ₂ H, pia ni monoprotiki kwa sababu tu atomu ya hidrojeni kutoka kundi la kaboksili (COOH) humenyuka kwa misingi:

Vile vile, besi za monoprotic ni besi ambazo zitakubali proton moja.

Asidi za diprotic zina atomi mbili za hidrojeni ionizable kwa molekuli; ionization ya asidi hizo hutokea katika hatua mbili. Ionization ya kwanza daima hufanyika kwa kiwango kikubwa kuliko ionization ya pili. Kwa mfano, asidi sulfuriki, asidi kali, ionizes kama ifuatavyo:

$$\begin{array}{}\\ \\ \text{Ionization ya kwanza:}{\text{H}}_2{\text{KWA HIVYO}}_4(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{HSO}}_4{}^{\text{-}}(aq)K_{\text{a}1}=\text{zaidi ya}{10}^{\text{2}}\text{; kujitenga kamili}\\ \text{Ionization ya pili:}{\text{HSO}}_4{}^{\text{-}}(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{KWA HIVYO}}_4{}^{\mathrm{2-}}(aq)K_{\text{a}2}=1.2\times {10}^{-2}\end{array}$$

Utaratibu huu wa ionization wa hatua kwa hatua hutokea kwa asidi zote za polyprotic. Asidi ya kaboni, H ₂ CO ₃, ni mfano wa asidi dhaifu ya diprotic. Ionization ya kwanza ya asidi ya kaboni huzalisha ions za hidroniamu na ions za bicarbonate kwa kiasi kidogo.

$$\begin{array}{l}\text{Ionization ya kwanza:}\\ {\text{H}}_2{\text{USHIRIKIANO}}_3(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{HCO}}_3{}^{\text{-}}(aq)K_{{\text{H}}_2{\text{USHIRIKIANO}}_3}=\frac{[{\text{H}}_3{\text{O}}^{\text{+}}][{\text{HCO}}_3{}^{\text{-}}]}{[{\text{H}}_2{\text{USHIRIKIANO}}_3]}=4.3\times {10}^{-7}\end{array}$$

Ioni ya bicarbonate pia inaweza kutenda kama asidi. Ni ionizes na hufanya ions hydronium na ions carbonate kwa kiasi kidogo hata.

$$\begin{array}{l}\text{Ionization ya pili:}\\ {\text{HCO}}_3{}^{\text{-}}(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{USHIRIKIANO}}_3{}^2(aq)K_{{\text{HCO}}_3{}^{\text{-}}}=\frac{[{\text{H}}_3{\text{O}}^{\text{+}}][{\text{USHIRIKIANO}}_3{}^2]}{[{\text{HCO}}_3{}^{\text{-}}]}=4.7\times {10}^{-11}\end{array}$$

$K_{{\text{H}}_2{\text{USHIRIKIANO}}_3}$ni kubwa kuliko$K_{{\text{HCO}}_3{}^{\text{-}}}$kwa sababu ya 10 ⁴, hivyo H ₂ CO ₃ ni mtayarishaji mkuu wa ion hydronium katika suluhisho. Hii ina maana kwamba kidogo ya${\text{HCO}}_3{}^{\text{-}}$iliyoundwa na ionization ya H ₂ CO ₃ ionizes kutoa ions hidronium (na ions carbonate), na viwango vya H ₃ O ⁺ na${\text{HCO}}_3{}^{\text{-}}$ni sawa sawa katika suluhisho safi la maji ya H ₂ CO ₃.

Kama kwanza ionization mara kwa mara ya asidi dhaifu diprotic ni kubwa kuliko ya pili kwa sababu ya angalau 20, ni sahihi kutibu ionization kwanza tofauti na mahesabu ya viwango kutokana na hayo kabla ya kuhesabu viwango vya aina kutokana na ionization baadae. Njia hii imeonyeshwa katika zoezi zifuatazo mfano.

Mfano 14.19

Ionization ya Asidi ya Diprotic

“Maji ya kaboni” yana kiasi cha kupendeza cha dioksidi kaboni iliyoharibika. Suluhisho ni tindikali kwa sababu CO ₂ humenyuka na maji ili kuunda asidi kaboni, H ₂ CO ₃. ni nini$[{\text{H}}_3{\text{O}}^{\text{+}}],$ $[{\text{HCO}}_3{}^{\text{-}}],$na$[{\text{USHIRIKIANO}}_3{}^2]$katika suluhisho iliyojaa ya CO ₂ na ya awali [H ₂ CO ₃] = 0.033 M?

$${\text{H}}_2{\text{USHIRIKIANO}}_3(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{HCO}}_3{}^{\text{-}}(aq)K_{\text{a}1}=4.3\times {10}^{-7}$$

$${\text{HCO}}_3{}^{\text{-}}(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{USHIRIKIANO}}_3{}^2(aq)K_{\text{a2}}=4.7\times {10}^{-11}$$

Suluhisho

Kama ilivyoonyeshwa na constants ionization, H ₂ CO ₃ ni asidi nguvu zaidi kuliko${\text{HCO}}_3{}^{\text{-}},$hivyo athari za ionization za stepwise zinaweza kutibiwa tofauti.

Mmenyuko wa kwanza wa ionization ni

$${\text{H}}_2{\text{USHIRIKIANO}}_3(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{HCO}}_3{}^{\text{-}}(aq)K_{\text{a}1}=4.3\times {10}^{-7}$$

Kutumia habari zinazotolewa, meza ya ICE kwa hatua hii ya kwanza imeandaliwa:

Kubadilisha viwango vya usawa katika equation ya usawa inatoa

$$K_{{\text{H}}_2{\text{USHIRIKIANO}}_3}=\frac{[{\text{H}}_3{\text{O}}^{\text{+}}][{\text{HCO}}_3{}^{\text{-}}]}{[{\text{H}}_2{\text{USHIRIKIANO}}_3]}=\frac{(x)(x)}{0.033-x}=4.3\times {10}^{-7}$$

Kutokana x <<<0.033 na kutatua mazao ya equation kilichorahisishwa

$$x=1.2\times {10}^{-4}$$

Jedwali la ICE linafafanuliwa x kama sawa na molarity ya ioni ya bicarbonate na molarity ya ioni ya hidronium:

$$[{\text{H}}_2{\text{USHIRIKIANO}}_3]=0.033M$$

$$[{\text{H}}_3{\text{O}}^{\text{+}}]=[{\text{HCO}}_3{}^{\text{-}}]=1.2\times {10}^{-4}M$$

Kutumia mkusanyiko wa ioni ya bicarbonate uliohesabiwa hapo juu, ionization ya pili inakabiliwa na hesabu sawa ya usawa:

$${\text{HCO}}_3{}^{\text{-}}(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{USHIRIKIANO}}_3{}^2(aq)$$

$$K_{{\text{HCO}}_3{}^{\text{-}}}=\frac{[{\text{H}}_3{\text{O}}^{\text{+}}][{\text{USHIRIKIANO}}_3{}^2]}{[{\text{HCO}}_3{}^{\text{-}}]}=\frac{(1.2\times {10}^{-4})[{\text{USHIRIKIANO}}_3{}^2]}{1.2\times {10}^{-4}}$$

$$[{\text{USHIRIKIANO}}_3{}^2]=\frac{(4.7\times {10}^{-11})(1.2\times {10}^{-4})}{1.2\times {10}^{-4}}=4.7\times {10}^{-11}M$$

Kwa muhtasari: katika usawa [H ₂ CO ₃] = 0.033 M;$[{\text{H}}_3{\text{O}}^{\text{+}}]$= 1.2$\times$10 ^-4;$[{\text{HCO}}_3{}^{\text{-}}]=1.2\times {10}^{-4}M;$ $[{\text{USHIRIKIANO}}_3{}^2]=4.7\times {10}^{-11}M.$

Angalia Kujifunza Yako

Mkusanyiko wa H ₂ S katika suluhisho la maji yenye maji yenye joto la kawaida ni takriban 0.1 M. Tumia$[{\text{H}}_3{\text{O}}^{\text{+}}],$[HS ^-], na [S ^2-] katika suluhisho:

$${\text{H}}_2\text{S}(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{INA}}^{\text{-}}(aq)K_{\text{a}1}=8.9\times {10}^{-8}$$

$${\text{INA}}^{\text{-}}(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{S}}^2(aq)K_{\text{a}2}=1.0\times {10}^{\mathrm{19-19}}$$

Jibu:

[H ₂ S] = 0.1 M;$[{\text{H}}_3{\text{O}}^{\text{+}}]$= [HS ^-] = 0.000094 M; [S ^2-] = 1$\times$10 ^-19 M

Asidi ya triprotiki ni asidi ambayo ina atomi tatu za ionizable H. Asidi ya fosforasi ni mfano mmoja:

$$\begin{array}{}\\ \text{Ionization ya kwanza:}{\text{H}}_3{\text{PO}}_4(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{H}}_2{\text{PO}}_4{}^{\text{-}}(aq)K_{\text{a}1}=7.5\times {10}^{\mathrm{1-3}}\\ \text{Ionization ya pili:}{\text{H}}_2{\text{PO}}_4{}^{\text{-}}(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{HPO}}_4{}^2(aq)K_{\text{a}2}=6.2\times {10}^{-8}\\ \text{Ionization ya tatu:}{\text{HPO}}_4{}^2(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{PO}}_4{}^{\mathrm{3-}}(aq)K_{\text{a}3}=4.2\times {10}^{-13}\end{array}$$

Kama kwa mifano ya asidi ya diprotic, kila mmenyuko wa ionization mfululizo ni mdogo zaidi kuliko wa zamani, unaojitokeza katika maadili ya kupungua kwa viwango vya asidi ya ionization ya hatua kwa hatua. Hii ni tabia ya jumla ya asidi ya poliprotic na mfululizo wa ionization mara nyingi hutofautiana na sababu ya 10 ⁵ hadi 10 ⁶.

Seti hii ya athari tatu za kujitenga inaweza kuonekana kufanya mahesabu ya viwango vya usawa katika suluhisho la H ₃ PO ₄ ngumu. Hata hivyo, kwa sababu mfululizo wa ionization mara kwa mara hutofautiana na sababu ya 10 ⁵ hadi 10 ⁶, tofauti kubwa zipo katika mabadiliko madogo katika mkusanyiko unaoongozana na athari za ionization. Hii inaruhusu matumizi ya mawazo na taratibu za hisabati, kama ilivyoonyeshwa katika mifano hapo juu.

Besi za poliprotiki zina uwezo wa kukubali ioni zaidi ya moja ya hidrojeni. Ioni ya carbonate ni mfano wa msingi wa diprotic, kwa sababu inaweza kukubali protoni mbili, kama inavyoonyeshwa hapa chini. Sawa na kesi ya asidi za poliprotic, angalia vipindi vya ionization hupungua kwa hatua ya ionization. Vivyo hivyo, mahesabu ya usawa yanayohusisha besi za poliprotiki hufuata mbinu sawa na zile za asidi za poliprotiki.

$$\begin{array}{}\\ \\ {\text{H}}_2\text{O}(l)+{\text{USHIRIKIANO}}_3{}^2(aq)\rightleftharpoons {\text{HCO}}_{\text{3}}{}^{-}(aq)+{\text{OH}}^{-}(aq)K_{\text{b1}}=2.1\times {10}^{-4}\\ {\text{H}}_2\text{O}(l)+H{{\text{USHIRIKIANO}}_3}^{-}(aq)\rightleftharpoons {\text{H}}_{\text{2}}{\text{USHIRIKIANO}}_{\text{3}}(aq)+{\text{OH}}^{-}(aq)K_{\text{B2}}=2.3\times {10}^{-8}\end{array}$$