22.14.14: Sura ya 14

1.

Mfano mmoja kwa NH ₃ kama asidi conjugate:${\text{NH}}_2{}^{\text{-}}+{\text{H}}^{\text{+}}⟶{\text{NH}}_3;$kama msingi wa conjugate:${\text{NH}}_4{}^{\text{+}}(aq)+{\text{OH}}^{\text{-}}(aq)⟶{\text{NH}}_3(aq)+{\text{H}}_2\text{O}(l)$

3.

(a)${\text{H}}_3{\text{O}}^{\text{+}}(aq)⟶{\text{H}}^{\text{+}}(aq)+{\text{H}}_2\text{O}(l);$(b)$\text{HCl}(\mathrm{aq})⟶{\text{H}}^{\text{+}}(aq)+{\text{Cl}}^{\text{-}}(aq);$(c)${\text{NH}}_3(aq)⟶{\text{H}}^{\text{+}}(aq)+{\text{NH}}_2{}^{\text{-}}(aq);$(d)${\text{CH}}_3{\text{USHIRIKIANO}}_2\text{H}(aq)⟶{\text{H}}^{\text{+}}(aq)+{\text{CH}}_3{\text{USHIRIKIANO}}_2{}^{\text{-}}(aq);$(e)${\text{NH}}_4{}^{\text{+}}(aq)⟶{\text{H}}^{\text{+}}(aq)+{\text{NH}}_3(aq);$(f)${\text{HSO}}_4{}^{\text{-}}(aq)⟶{\text{H}}^{\mathrm{+}}(aq)+{\text{KWA HIVYO}}_4{}^{\text{2-}}(aq)$

5.

(a)${\text{H}}_2\text{O}(l)+{\text{H}}^{\text{+}}(aq)⟶{\text{H}}_3{\text{O}}^{\text{+}}(aq);$(b)${\text{OH}}^{\text{-}}(aq)+{\text{H}}^{\text{+}}(aq)⟶{\text{H}}_2\text{O}(l);$(c)${\text{NH}}_3(aq)+{\text{H}}^{\text{+}}(aq)⟶{\text{NH}}_4{}^{\text{+}}(aq);$(d)${\text{CN}}^{\text{-}}(aq)+{\text{H}}^{\text{+}}(aq)⟶\text{HCN}(aq);$(e)${\text{S}}^{\mathrm{2-}}(aq)+{\text{H}}^{\text{+}}(aq)⟶{\text{INA}}^{\text{-}}(aq);$(f)${\text{H}}_2{\text{PO}}_4{}^{\text{-}}(aq)+{\text{H}}^{\text{+}}(aq)⟶{\text{H}}_3{\text{PO}}_4(aq)$

7.

(a) H ₂ O, O ^2; (b) H ₃ O ⁺, OH ^-; (c) H ₂ CO _3,${\text{USHIRIKIANO}}_3{}^{\text{2-}};$(d)${\text{NH}}_4{}^{\text{+}},$ ${\text{NH}}_2{}^{\text{-}};$(e) H ₂ SO _4,${\text{KWA HIVYO}}_4{}^{\text{2-}};$(f)${\text{H}}_3{\text{O}}_2{}^{\text{+}},$ ${\text{HO}}_2{}^{\text{-}};$(g) H ₂ S; S ²; (h)${\text{H}}_6{\text{N}}_2{}^{\mathrm{2+}},$H ₄ N ₂

9.

Maandiko ni asidi ya Brønsted-Lowry = BA; msingi wake wa conjugate = CB; Msingi wa Brønsted-Lowry = BB; asidi yake ya conjugate = CA. (a) HNO ₃ (BA), H ₂ O (BB), H ₃ O ⁺ (CA),${\text{HAPANA}}_3{}^{\text{-}}(\text{CB});$(b) CN ^- (BB), H ₂ O (BA), HCN (CA), OH ^- (CB); (c) H ₂ SO ₄ (BA), Cl ^- (BB), HCl (CA),${\text{HSO}}_4{}^{\text{-}}(\text{CB});$(d)${\text{HSO}}_4{}^{\text{-}}(\text{BA}),$OH ^- (BB),${\text{KWA HIVYO}}_4{}^{\text{2-}}$(CB), H ₂ O (CA); (e) O ² (BB), H ₂ O (BA) OH ^- (CB na CA); (f) [Cu (H ₂ O) ₃ (OH)] ⁺ (BB), [Al (H ₂ O) ₆] ³⁺ (BA), [Cu (H ₂ O) ₄] ²⁺ (CA), [Al (H) ₂ O) ₅ (OH)] ²⁺ (CB); (g) H ₂ S (BA),${\text{NH}}_2{}^{\text{-}}(\text{BB}),$HS ^- (CB), NH ₃ (CA)

11.

Spishi za amphiprotiki zinaweza ama kupata au kupoteza protoni katika mmenyuko wa kemikali, hivyo kutenda kama msingi au asidi. Mfano ni H ₂ O. kama asidi: ${\text{H}}_2\text{O}(aq)+{\text{NH}}_3(aq)\rightleftharpoons {\text{NH}}_4{}^{\text{+}}(aq)+{\text{OH}}^{\text{-}}(aq).$Kama msingi:${\text{H}}_2\text{O}(aq)+\text{HCl}(aq)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{Cl}}^{\text{-}}(aq)$

13.

amphiprotic: (a)${\text{NH}}_3+{\text{H}}_3{\text{O}}^{\text{+}}⟶{\text{NH}}_4\text{OH}+{\text{H}}_2\text{O},$ ${\text{NH}}_3+{\text{OCH}}_3{}^{\text{-}}⟶{\text{NH}}_2{}^{\text{-}}+{\text{CH}}_3\text{OH};$(b)${\text{HPO}}_4{}^{\text{2-}}+{\text{OH}}^{\text{-}}⟶{\text{PO}}_4{}^{\text{3-}}+{\text{H}}_2\text{O},$ ${\text{HPO}}_4{}^{\text{2-}}+{\text{Chlo}}_4⟶{\text{H}}_2{\text{PO}}_4{}^{\text{-}}+{\text{ClO}}_4{}^{\text{-}};$si amphiprotic: (c) Br ^-; (d)${\text{NH}}_4{}^{\text{+}};$(e)${\text{Aso}}_4{}^{\text{3-}}$

15.

Katika suluhisho la neutral [H ₃ O ⁺] = [OH ^-]. Saa 40 °C, [H ₃ O ⁺] = [^{OH -}] = (2.910 × 10 ^-14) ^1/2 = 1.7$\times$^10-7.

17.

x = 3.051$\times$10 ^-7 M = [H ₃ O ⁺] = [OH ^-]; pH = -log3.051$\times$10 ^-7 = - (-6.5156) = 6.5156; poH = pH = 6.5156

19.

(a) pH = 3.587; poH = 10.413; (b) PoH = 0.68; pH = 13.32; (c) PoH = 3.85; pH = 10.15; (d) PoH = -0.40; pH = 14.4

21.

[H ₃ O ⁺] = 3.0$\times$10 ^-7 M; [OH ^-] = 3.3$\times$10 ^-8 M

23.

[H ₃ O ⁺] = 1$\times$10 ^-2 M; [OH ^∙] = 1$\times$10 ^-12 M

25.

[OH ^-] = 3.1$\times$10 ^-12 M

27.

Chumvi ionizes katika suluhisho, lakini anion inachukua kidogo na maji ili kuunda asidi dhaifu. Tabia hii pia ^{huunda OH -}, ambayo husababisha suluhisho kuwa msingi.

29.

[H ₂ O] > [CH ₃ CO ₂ H] >$[{\text{H}}_3{\text{O}}^{\text{+}}]$≈$[{\text{CH}}_3{\text{USHIRIKIANO}}_2{}^{\text{-}}]$> [OH ^-]

31.

Hali ya oxidation ya sulfuri katika H ₂ SO ₄ ni kubwa kuliko hali ya oxidation ya sulfuri katika H ₂ SO ₃.

33.

$\begin{array}{cccccc}\text{Mg}{(\text{OH})}_2(s)+& \text{2hCl}(aq)& ⟶& {\text{Mg}}^{\mathrm{2+}}(aq)+& 2{\text{Cl}}^{\text{-}}(aq)+& {\text{2H}}_2\text{O}(l)\\ \text{BB}& \text{BA}& & & \text{CB}& \text{CA}& \end{array}$

35.

$K_{\text{a}}=2.3\times {10}^{-11}$

37.

Msingi wa nguvu au asidi kali ni moja na K kubwa au _{K a, kwa} mtiririko huo. Katika mifano hii miwili, wao ni (CH ₃) ₂ NH na${\text{CH}}_3{\text{NH}}_3{}^{\text{+}}.$

39.

triethylamine

41.

(a)${\text{HSO}}_4{}^{\text{-}};$electronegativity ya juu ya ion kuu. (b) H ₂ O; NH ₃ ni msingi na maji ni neutral, au kuamua kwa misingi _ya K maadili. (c) HI; PH ₃ ni dhaifu kuliko HCl; HCl ni dhaifu kuliko HI. Hivyo, PH ₃ ni dhaifu kuliko HI. (d) PH ₃; katika misombo ya binary ya hidrojeni na nonmetals, asidi huongezeka kwa kipengele cha chini katika kikundi. (e) HbR; kwa kipindi, asidi huongezeka kutoka kushoto kwenda kulia; katika kikundi, huongezeka kutoka juu hadi chini. Br ni upande wa kushoto na chini ya S, hivyo HbR ni asidi kali.

43.

(a) NaHSEO ₃ <NaHSO ₃ <NaHSO ₄; katika asidi ya polyoxy, kipengele cha kati cha electronegative zaidi, katika kesi hii-huunda asidi yenye nguvu. Idadi kubwa ya atomi za oksijeni kwenye atomi ya kati (ikitoa hali ya juu ya oxidation) pia inajenga kutolewa zaidi kwa atomi za hidrojeni, na kusababisha asidi kali. Kama chumvi, asidi huongezeka kwa namna ile ile. (b)${\text{ClO}}_2{}^{\text{-}}<{\text{Bro}}_2{}^{\text{-}}<{\text{IO}}_2{}^{\text{-}};$msingi wa anions katika mfululizo wa asidi itakuwa kinyume cha asidi katika oxyacids yao. Asidi huongezeka kama electronegativity ya atomi kuu huongezeka. Cl ni electronegative zaidi ya Br, na mimi ni electronegative angalau ya tatu. (c) HOI <HOBR <HoCl; katika mfululizo wa aina moja ya oksiasidi, asidi huongezeka kadiri electronegativity ya atomi ya kati inavyoongezeka. Cl ni electronegative zaidi ya Br, na mimi ni electronegative angalau ya tatu. (d) HoCl <HoClO <HoClo ₂ <HoClO ₃; katika mfululizo wa oksijeni ya elementi moja ya kati, asidi huongezeka kadiri idadi ya atomi za oksijeni inavyoongezeka (au kama hali ya oksidi ya atomi ya kati inavyoongezeka). (e)${\text{Ya}}^{\text{-}}<{\text{INA}}^{\text{-}}<<<{\text{PH}}_2{}^{\text{-}}<{\text{NH}}_2{}^{\text{-}};$ ${\text{PH}}_2{}^{\text{-}}$na${\text{NH}}_2{}^{\text{-}}$ni anions ya besi dhaifu, hivyo hufanya kama besi kali kuelekea H ⁺.${\text{Ya}}^{\text{-}}$na HS ^- ni anions ya asidi dhaifu, hivyo wana tabia ya chini ya msingi. Katika kundi la mara kwa mara, kipengele cha electronegative zaidi kina anion ya msingi zaidi. (f)${\text{Bro}}_4{}^{\text{-}}<{\text{Bro}}_3{}^{\text{-}}<{\text{Bro}}_2{}^{\text{-}}<{\text{Bro}}^{\text{-}};$na idadi kubwa ya atomi za oksijeni (yaani, kama hali ya oxidation ya ioni kuu inavyoongezeka), asidi inayofanana inakuwa tindikali zaidi na anion hivyo chini ya msingi.

45.

$[{\text{H}}_2\text{O}]>[{\text{C}}_6{\text{H}}_4\text{OH}({\text{USHIRIKIANO}}_2\text{H})]>{\text{[H}}^{\text{+}}\text{]}\text{0}>{\text{[C}}_6{\text{H}}_4\text{OH}{({\text{USHIRIKIANO}}_2)}^{\text{-}}]\gg [{\text{C}}_6{\text{H}}_4\text{O}{({\text{USHIRIKIANO}}_2\text{H})}^{\text{-}}]>[{\text{OH}}^{\text{-}}]$

47.

1. Fikiria kwamba mabadiliko katika mkusanyiko wa awali wa asidi kama usawa umeanzishwa yanaweza kupuuzwa, hivyo mkusanyiko huu unaweza kudhaniwa mara kwa mara na sawa na thamani ya awali ya mkusanyiko wa asidi ya jumla. Tuseme tunaweza kupuuza mchango wa maji kwa mkusanyiko wa usawa wa H ₃ O ⁺.

48.

(b) Kuongezea kwa HCl

50.

(a) Kuongeza HCl itaongeza H ₃ O ⁺ ions, ambayo kisha kuguswa na OH ^{- ions}, kupunguza mkusanyiko wao. Msawazo utabadilika kwa haki, kuongeza mkusanyiko wa HNO ₂, na kupunguza ukolezi wa${\text{HAPANA}}_2{}^{\text{-}}$ioni. (b) Kuongeza HNO ₂ huongeza mkusanyiko wa HNO ₂ na mabadiliko ya usawa kwa upande wa kushoto, na kuongeza mkusanyiko wa${\text{HAPANA}}_2{}^{\text{-}}$ions na kupungua kwa mkusanyiko wa ions ^{OH -}. (c) Kuongeza NaOH anaongeza ^{OH -} ions, ambayo mabadiliko ya usawa kwa upande wa kushoto, kuongeza mkusanyiko wa${\text{HAPANA}}_2{}^{\text{-}}$ions na kupungua kwa viwango vya HNO ₂. (d) Kuongeza NaCl haina athari kwa viwango vya ions. (e) Kuongeza KNO ₂ anaongeza${\text{HAPANA}}_2{}^{\text{-}}$ions na mabadiliko ya usawa kwa haki, kuongeza viwango vya HNO ₂ na ^{OH -} ion.

52.

Hii ni kesi ambayo suluhisho ina mchanganyiko wa asidi ya nguvu tofauti za ionization. Katika suluhisho, HCO ₂ H ipo hasa kama molekuli ya HCO ₂ H kwa sababu ionization ya asidi dhaifu inakandamizwa na asidi kali. Kwa hiyo, HCO ₂ H inachangia kiasi kidogo cha ions hydronium kwa suluhisho. Asidi yenye nguvu, HCl, ni mtayarishaji mkuu wa ions za hidronium kwa sababu ni ionized kabisa. Katika suluhisho hilo, asidi yenye nguvu huamua mkusanyiko wa ioni za hidronium, na ionization ya asidi dhaifu ni fasta na [H ₃ O ⁺] zinazozalishwa na asidi kali.

54.

(a)$K_{\text{b}}=1.8\times {10}^{-5};$ (b)$K_{\text{a}}=4.5\times {10}^{-4};$ (c)$K_{\text{b}}=6.4\times {10}^{-5};$ (d)$K_{\text{a}}=5.6\times {10}^{-10}$

56.

$K_{\text{a}}=1.2\times {10}^{-2}$

58.

(a)$K_{\text{b}}=4.3\times {10}^{-12}$ (b)$K_{\text{a}}=1.6\times {10}^{10}$ (c)$K_{\text{b}}=5.9\times {10}^8$ (d)$K_{\text{b}}=4.2\times {10}^{\mathrm{1-3}}$ (e)$K_{\text{b}}=2.3\times {10}^5$ (f)$K_{\text{b}}=6.3\times {10}^{-13}$

60.

(a)$\frac{[{\text{H}}_3{\text{O}}^{\text{+}}][{\text{ClO}}^{\text{-}}]}{[\text{Chlo}]}=\frac{(x)(x)}{(0.0092-x)}\approx \frac{(x)(x)}{0.0092}=2.9\times {10}^{-8}$
Kutatua kwa x inatoa 1.63$\times$10 ^-5 M. Thamani hii ni chini ya 5% ya 0.0092, hivyo dhana kwamba inaweza kupuuzwa ni halali. Hivyo, viwango vya spishi za solute kwenye usawa ni:
[H ₃ O ⁺] = [ClO ^—] = 1.6$\times$10 ^-5 M
[HClo ^—] = 0.0092 M
[OH ^∙] = 6.1$\times$10 ^-10 M;
(b)$\frac{[{\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[{\text{C}}_6{\text{H}}_5{\text{NH}}_2]}=\frac{(x)(x)}{(0.0784-x)}\approx \frac{(x)(x)}{0.0784}=4.3\times {10}^{-10}$
Kutatua kwa x inatoa 5.81$\times$10 ^-6 M. Thamani hii ni chini ya 5% ya 0.0784, hivyo dhana kwamba inaweza kupuuzwa ni halali. Hivyo, viwango vya aina za solute katika usawa ni:
$[{\text{C}}_6\text{H}5{\text{NH}}_3{}^{\text{+}}]$= [OH ^-] = 5.8$\times$10 ^-6 M
[C ₆ H ₅ NH ₂] = 0.0784 M
[H ₃ O ⁺] = 1.7$\times$10 ^-9 M;
(c)$\frac{[{\text{H}}_3{\text{O}}^{\text{+}}][{\text{CN}}^{\text{-}}]}{[\text{HCN}]}=\frac{(x)(x)}{(0.0810-x)}\approx \frac{(x)(x)}{0.0810}=4.9\times {10}^{-10}$
Kutatua kwa x anatoa 6.30$\times$10 ^-6 M. Thamani hii ni chini ya 5% ya 0.0810, hivyo dhana kwamba inaweza kupuuzwa ni halali. Hivyo, viwango vya spishi za solute kwenye usawa ni:
[H ₃ O ⁺] = [CN ^∙] = 6.3$\times$10 ^-6 M
[HCN] = 0.0810 M
[OH ^∙] = 1.6$\times$10 ^-9 M;
(d)$\frac{[{({\text{CH}}_3)}_3{\text{NH}}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[{({\text{CH}}_3)}_3\text{N}]}=\frac{(x)(x)}{(0.11-x)}\approx \frac{(x)(x)}{0.11}=6.3\times {10}^{-5}$
Kutatua kwa x inatoa 2.63$\times$10 ^-3 M. Thamani hii ni chini ya 5% ya 0.11, hivyo dhana kwamba inaweza kupuuzwa ni halali. Hivyo, viwango vya spishi za solute kwenye usawa ni:
[(CH ₃) ₃ NH ⁺] = [OH ^-] = 2.6$\times$10 ^1-3 M
[(CH _{3) 3} N] = 0.11 M
[H ₃ O ⁺] = 3.8$\times$10 ^-12 M;
(e)$\frac{[\text{Fe}{({\text{H}}_2\text{O})}_5({\text{OH})}^{\text{+}}][{\text{H}}_3{\text{O}}^{\text{+}}]}{[\text{Fe}{({\text{H}}_2\text{O})}_6{}^{\mathrm{2+}}]}=\frac{(x)(x)}{(0.120-x)}\approx \frac{(x)(x)}{0.120}=1.6\times {10}^{-7}$
Kutatua kwa x inatoa 1.39$\times$10 ^-4 M. Thamani hii ni chini ya 5% ya 0.120, hivyo dhana kwamba inaweza kupuuzwa ni halali. Hivyo, viwango vya spishi za solute katika usawa ni:
[Fe (H ₂ O) ₅ (OH) ⁺] = [H ₃ O ⁺] = 1.4$\times$10 ^-4 M
$[\text{Fe}{({\text{H}}_2\text{O})}_6{}^{\mathrm{2+}}]$= 0.120 M
[OH ^-] = 7.2$\times$10 ^-11 M

62.

pH = 2.41

64.

[C ₁₀ H ₁₄ N ₂] = 0.049 M; [C ₁₀ H ₁₄ N ₂ H ⁺] = 1.9$\times$10 ^-4 M; $[{\text{C}}_{10}{\text{H}}_{14}{\text{N}}_2{\text{H}}_2{}^{\mathrm{2+}}]$= 1.4$\times$10 ^-11 M; [OH ^-] = 1.9$\times$10 ^-4 M; [H ₃ O ⁺] = 5.3$\times$10 ^-11 M

66.

$K_{\text{a}}=1.2\times {10}^{-2}$

68.

$K_{\text{b}}=1.77\times {10}^{-5}$

70.

(a) tindikali; (b) msingi; (c) tindikali; (d) neutral

72.

[H ₃ O ⁺] na$[{\text{HCO}}_3{}^{\text{-}}]$ni kivitendo sawa

74.

[C ₆ H ₄ (CO ₂ H) ₂] 7.2$\times$10 ^1-3 M, [C ₆ H ₄ (CO ₂ H) (CO ₂) ^-] = [H ₃ O ⁺] 2.8$\times$10 ^-3 M,$[{\text{C}}_6{\text{H}}_4{({\text{USHIRIKIANO}}_2)}_2{}^2]$3.9$\times$10 ^-6 M, [OH ∙^] 3.6$\times$10 ^-12 M

76.

(a)$K_{\text{a2}}=1.5\times {10}^{-11};$
(b)$K_{\text{b}}=4.3\times {10}^{-12};$
(c)$\frac{[{\text{Te}}^2][{\text{H}}_3{\text{O}}^{\text{+}}]}{[{\text{The}}^{\text{-}}]}=\frac{(x)(0.0141+x)}{(0.0141-x)}\approx \frac{(x)(0.0141)}{0.0141}=1.5\text{×}{10}^{-11}$
Kutatua kwa x hutoa 1.5$\times$10 ^-11 M. Kwa hiyo, ikilinganishwa na 0.014 M, thamani hii ni duni (1.1$\times$10 ^-7%).

78.

Zaidi ya H ₃ O ⁺ huondolewa hasa na majibu: ${\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{H}}_2{\text{PO}}_4{}^{\text{-}}(aq)⟶{\text{H}}_3{\text{PO}}_4(aq)+{\text{H}}_2\text{O}(l)$
Msingi wa ziada huondolewa na majibu: ${\text{OH}}^{\text{-}}(aq)+{\text{H}}_3{\text{PO}}_4(aq)⟶{\text{H}}_2{\text{PO}}_4{}^{\text{-}}(aq)+{\text{H}}_2\text{O}(l)$

80.

[H ₃ O ⁺] = 1.5$\times$10 ^-4 M

82.

[OH ^-] = 4.2$\times$10 ^-4 M

84.

(a) HCl iliyoongezwa itaongeza mkusanyiko wa H ₃ O ⁺ kidogo, ambayo itachukua hatua${\text{CH}}_3{\text{USHIRIKIANO}}_2{}^{\text{-}}$na kuzalisha CH ₃ CO ₂ H katika mchakato. Hivyo,$[{\text{CH}}_3{\text{USHIRIKIANO}}_2{}^{\text{-}}]$itapungua na [CH ₃ CO ₂ H] huongezeka. (b) aliongeza KCH ₃ CO ₂ itaongeza mkusanyiko wa$[{\text{CH}}_3{\text{USHIRIKIANO}}_2{}^{\text{-}}]$ambayo itachukua hatua na H ₃ O ⁺ na kuzalisha CH ₃ CO ₂ H katika mchakato. Hivyo, [H ₃ O ⁺] hupungua kidogo na [CH ₃ CO ₂ H] huongezeka. (c) NaCl iliyoongezwa haitakuwa na athari kwenye mkusanyiko wa ions. (d) KOH iliyoongezwa itazalisha ^ions ya OH, ambayo itachukua hatua na H ₃ O ⁺, hivyo kupunguza [H ₃ O ⁺]. Baadhi ya ziada CH ₃ CO ₂ H itakuwa dissociate, kuzalisha$[{\text{CH}}_3{\text{USHIRIKIANO}}_2{}^{\text{-}}]$ions katika mchakato. Hivyo, [CH ₃ CO ₂ H] hupungua kidogo na$[{\text{CH}}_3{\text{USHIRIKIANO}}_2{}^{\text{-}}]$huongezeka. (e) CH ₃ CO ₂ H iliyoongezwa itaongeza mkusanyiko wake, na kusababisha zaidi ya kujitenga na kuzalisha zaidi$[{\text{CH}}_3{\text{USHIRIKIANO}}_2{}^{\text{-}}]$na H ₃ O ⁺ katika mchakato. Hivyo, [H ₃ O ⁺] huongezeka kidogo na$[{\text{CH}}_3{\text{USHIRIKIANO}}_2{}^{\text{-}}]$huongezeka.

86.

pH = 8.95

88.

37 g (0.27 mol)

90.

(a) pH = 5.222; (b) Suluhisho ni tindikali. (c) pH = 5.220

92.

Katika hatua ya ulinganifu katika titration ya msingi dhaifu na asidi kali, suluhisho la kusababisha ni tindikali kidogo kutokana na kuwepo kwa asidi conjugate. Kwa hiyo, chagua kiashiria kinachobadilisha rangi katika aina ya tindikali na mabano ya pH kwenye hatua ya ulinganifu. Methyl machungwa ni mfano mzuri.

94.

(a) pH = 2.50; (b) pH = 4.01; (c) pH = 5.60; (d) pH = 8.35; (e) pH = 11.08