Skip to main content
Global

15.E: Msawazo wa Madarasa mengine ya Majibu (Mazoezi)

  1. Last updated
  2. Save as PDF
  • Page ID
    176763

    • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    15.1: Upepo na Uharibifu

    Q15.1.1

    Jaza mabadiliko katika viwango kwa kila moja ya athari zifuatazo:

    \ (\ kuanza {alignat} {3}
    &\ ce {AGI} (s) &\ ce {Ag+} (aq)\, +\, && ce {I-} (aq)\
    & &&x &\ kusisitiza {\ hnafasi {45px}}
    \ mwisho {alignat}\)

    \ (\ kuanza {alignat} {3}
    &\ ce {CaCO3} (s) &\ ce {Ca ^ 2+} (aq)\, +\, &\ ce {CO3 ^ 2-} (aq)\\
    & &&\ kusisitiza {\ hnafasi {45px}}} &x
    \ mwisho {alignat}\)

    \ (\ kuanza {alignat} {3}
    &\ ce {Mg (OH) 2} (s) &\ ce {Mg ^ 2+} (aq)\, +\, &\ ce {2OH-} (aq)\\
    & &&x &\ kusisitiza {\ hnafasi {45px}}
    \ mwisho {alignat}\)

    \ (\ kuanza {alignat} {3}
    &\ ce {Mg3 (PO4) 2} (s) &\ ce {3Mg ^ 2+} (aq)\, +\, &\ ce {2PO4 ^ 3-} (aq)\\
    & &&&x\ kusisitiza {\ hnafasi {45px}}
    \ mwisho {alignat}\)

    \ (kuanza {alignat} {3}
    &\ ce {Ca5 (PO4) 3OH} (s) &\ ce {5Ca^2+} (aq)\, +\, &\ ce {3PO4 ^ 3-} (aq)\, +\, &\ ce {OH-} (aq)\ & & &\ kusisitiza {\ hnafasi {45px}} &\ kusisitiza {\ hspace {45px}}
    &\ kusisitiza {\ px} &\ kusisitiza {\ hspace {45px}} &\ hspace {45px}} &&x
    \ mwisho {alignat}\)

    S15.1.1

    \ (\ kuanza {alignat} {3}
    &\ ce {AGI} (s) &&\ ce {Ag+} (aq)\, +\, &&\ ce {I-} (aq)\\
    & &&x &&\ kusisitiza {x}
    \ mwisho {alignat}\)

    \ (\ kuanza {alignat} {3}
    &\ ce {CaCO3} (s) &&\ ce {Ca^2+} (aq)\, +\, &&\ ce {CO3 ^ 2-} (aq)\\
    & &&&\ kusisitiza {x} &&x
    \ mwisho {alignat}\)

    \ (\ kuanza {alignat} {3}
    &\ ce {Mg (OH) 2} (s) &&\ ce {Mg ^ 2+} (aq)\, +\, &&\ ce {2OH-} (aq)\\
    & &&x &\ kusisitiza {2x}
    \ mwisho {alignat}\)

    \ (\ kuanza {alignat} {3}
    &\ ce {Mg3 (PO4) 2} (s) &&\ ce {3Mg ^ 2+} (aq)\, +\, &&\ ce {2PO4 ^ 3-} (aq)\\
    & &&&\ kusisitiza {3x} &2x
    \ mwisho {alignat}\)

    \ (\ kuanza {alignat} {3}
    &\ ce {Ca5 (PO4) 3OH} (s) &&\ ce {5Ca^2+} (aq)\, +\, &&\ ce {3PO4 ^ 3-} (aq)\, +\, &\ ce {OH-} (aq)\\ & & & kusisitiza {5x}
    &\ kusisitiza {3x} &&\ kusisitiza {3x} && x
    \ mwisho {alignat}\)

    Q15.1.2

    Jaza mabadiliko katika viwango kwa kila moja ya athari zifuatazo:

    \ (\ kuanza {alignat} {3}
    &\ ce {BaSO4} (s) &\ ce {ba ^ 2+} (aq)\, +\, &\ ce {SO4 ^ 2-} (aq)\\
    & &&x &\ kusisitiza {\ hnafasi {45px}}
    \ mwisho {alignat}\)

    \ (\ kuanza {alignat} {3}
    &\ ce {Ag2SO4} (s) &&\ ce {2Ag +} (aq)\, +\, &\ ce {SO4 ^ 2-} (aq)\\
    & &&\ kusisitiza {\ hnafasi {45px}}} &x
    \ mwisho {alignat}\)

    \ (\ kuanza {alignat} {3}
    &\ ce {Al (OH) 3} (s) &&\ ce {Al^3+} (aq)\, +\, && ce {3OH-} (aq)\\
    & &&x &\ kusisitiza {\ hnafasi {45px}}
    \ mwisho {alignat}\)

    \ (kuanza {alignat} {3}
    &\ ce {Pb (OH) Cl} (s) &&\ ce {Pb ^ 2+} (aq)\, +\, &\ ce {OH-} (aq)\, +\,
    && ce {Cl-} (aq)\ & & & & & kusisitiza {\ hnafasi {45px}}
    \ mwisho {alignat}\)

    \ (\ kuanza {alignat} {3}
    &\ ce {Ca3 (aSO4) 2} (s) &&\ ce {3Ca^2+} (aq)\, +\, &&\ ce {2aSO4 ^ 3-} (aq)\\
    & &&& 3x &\ kusisitiza {\ hnafasi {45px}}
    \ mwisho {alignat}\)

    Q15.1.3

    Je, viwango vya Ag + na\(\ce{CrO4^2-}\) katika suluhisho iliyojaa juu ya 1.0 g ya Ag 2 CrO 4 imara hubadilika wakati 100 g ya Ag 2 CrO 4 imara inaongezwa kwenye mfumo? Eleza.

    S15.1.3

    Hakuna mabadiliko. Mango ina shughuli ya 1 ikiwa kuna kidogo au mengi.

    Q15.1.4

    Je, viwango vya Pb 2+ na S 2— hubadilishaje wakati K 2 S inapoongezwa kwenye ufumbuzi uliojaa wa PBs?

    Q15.1.5

    Ni maelezo gani ya ziada tunahitaji kujibu swali linalofuata: Je, usawa wa bromidi ya fedha imara na ufumbuzi uliojaa wa ions zake huathiriwa wakati joto linafufuliwa?

    S15.1.5

    Umumunyifu wa bromidi ya fedha kwenye joto jipya lazima lijulikane. Kwa kawaida umumunyifu huongezeka na baadhi ya bromidi ya fedha imara itafuta.

    Q15.1.6

    Ni ipi kati ya misombo yafuatayo kidogo ya mumunyifu ina umumunyifu mkubwa kuliko ile iliyohesabiwa kutoka kwa bidhaa yake ya umumunyifu kwa sababu ya hidrolisisi ya sasa ya anion: CoSO 3, CuI, PBCl 3, PBCl 2, Tl 2 S, kClo 4?

    Q15.1.7

    Ni ipi kati ya misombo yafuatayo ya mumunyifu kidogo ina umumunyifu mkubwa kuliko ile iliyohesabiwa kutoka kwa bidhaa yake ya umumunyifu kwa sababu ya hidrolisisi ya sasa ya anion: AgCl, BaSO 4, CaF 2, Hg 2 I 2, MnCo 3, ZnS, PBs?

    S15.1.7

    CaF 2, MnCo 3, na ZNs

    Q15.1.8

    Andika equation ionic kwa ajili ya kufutwa na bidhaa umumunyifu (K sp) kujieleza kwa kila moja ya yafuatayo misombo kidogo mumunyifu ionic:

    1. PbCl 2
    2. Umri 2 S
    3. Sr 3 (PRO 42)
    4. SRSO 4

    Q15.1.9

    Andika equation ionic kwa ajili ya kufutwa na kujieleza K sp kwa kila moja ya yafuatayo misombo kidogo mumunyifu ionic:

    1. Laf 3
    2. caco 3
    3. Umri 2 SEO - 4
    4. Pb (OH) 2

    Q15.1.10

    1. \(\ce{LaF3}(s)⇌\ce{La^3+}(aq)+\ce{3F-}(aq) \hspace{20px} K_\ce{sp}=\ce{[La^3+][F- ]^3};\)
    2. \(\ce{CaCO3}(s)⇌\ce{Ca^2+}(aq)+\ce{CO3^2-}(aq) \hspace{20px} K_\ce{sp}=\ce{[Ca^2+][CO3^2- ]};\)
    3. \(\ce{Ag2SO4}(s)⇌\ce{2Ag+}(aq)+\ce{SO4^2-}(aq) \hspace{20px} K_\ce{sp}=\ce{[Ag+]^2[SO4^2- ]};\)
    4. \(\ce{Pb(OH)2}(s)⇌\ce{Pb^2+}(aq)+\ce{2OH-}(aq) \hspace{20px} K_\ce{sp}=\ce{[Pb^2+][OH- ]^2}\)

    Q15.1.11

    Kitabu cha Kemia na Fizikia hutoa umumunyifu wa misombo ifuatayo kwa gramu kwa 100 ml ya maji. Kwa sababu misombo hii ni kidogo tu mumunyifu, kudhani kwamba kiasi haibadilika juu ya kufutwa na kuhesabu bidhaa umumunyifu kwa kila mmoja.

    1. BasiF 6, 0.026 g/100 ml (ina\(\ce{SiF6^2-}\) ions)
    2. Ce (IO 3) 4, 1.5 × 10 —2 g/100 ml
    3. Gd 2 (SO 4) 3, 3.98 g/100 ml
    4. (NH 4) 2 PTBR 6, 0.59 g/100 ml (ina\(\ce{PtBr6^2-}\) ions)

    Q15.1.12

    Kitabu cha Kemia na Fizikia hutoa umumunyifu wa misombo ifuatayo kwa gramu kwa 100 ml ya maji. Kwa sababu misombo hii ni kidogo tu mumunyifu, kudhani kwamba kiasi haibadilika juu ya kufutwa na kuhesabu bidhaa umumunyifu kwa kila mmoja.

    1. BaseO 4, 0.0118 g/100 ml
    2. Ba (BRO 3) 2 •H 2 O, 0.30 g/100 ml
    3. NH 4 MGAO 4 • 6H 2 O, 0.038 g/100 ml
    4. La 2 (MoO 4) 3, 0.00179 g/100 ml

    S15.1.12

    (a) 1.77 × 10 —7; 1.6 × 10 —6; 2.2 × 10 —9; 7.91 × 10 —22

    Q15.1.13

    Tumia bidhaa za umumunyifu na utabiri ni ipi ya chumvi zifuatazo ni mumunyifu zaidi, kwa suala la moles kwa lita, katika maji safi: CaF 2, Hg 2 Cl 2, PBi 2, au Sn (OH) 2.

    Q15.1.14

    Kwa kuzingatia kwamba hakuna usawa isipokuwa uharibifu unaohusika, uhesabu umumunyifu wa molar wa kila moja ya yafuatayo kutoka kwa bidhaa zake za umumunyifu:

    1. KHC 4 H 4 O 6
    2. PBi 2
    3. Ag 4 [Fe (CN) 6], chumvi zenye\(\ce{Fe(CN)4-}\) chuma
    4. Hg 2 I 2

    S15.1.15

    2 × 10 —2 M; 1.3 × 10 —3 M; 2.27 × 10 —9 M; 2.2 × 10 —10 M

    Q15.1.16

    Kwa kuzingatia kwamba hakuna usawa isipokuwa uharibifu unaohusika, uhesabu umumunyifu wa molar wa kila moja ya yafuatayo kutoka kwa bidhaa zake za umumunyifu:

    1. Umri 2 SEO - 4
    2. Pbbr 2
    3. AGI
    4. CaC 2 O 4 • H 2 O

    Q15.1.X

    Kwa kuzingatia kwamba hakuna usawa isipokuwa kuvunjwa unahusishwa, uhesabu mkusanyiko wa aina zote za solute katika kila moja ya ufumbuzi wafuatayo wa chumvi unaowasiliana na suluhisho iliyo na ion ya kawaida. Onyesha kwamba mabadiliko katika viwango vya awali vya ions ya kawaida yanaweza kupuuzwa.

    1. GCl (s) katika 0.025 M NaCl
    2. CaF 2 (s) katika 0.00133 M KF
    3. Ag 2 SO 4 (s) katika 0.500 L ya suluhisho iliyo na 19.50 g ya K 2 SO 4
    4. Zn (OH) 2 (s) katika suluhisho iliyopigwa kwa pH ya 11.45

    S15.1.X

    7.2 × 10 -9 M = [Ag +], [Cl ∙] = 0.025 M

    Angalia:\(\dfrac{7.2×10^{−9}\:M}{0.025\:M}×100\%=2.9×10^{−5}\%\), mabadiliko yasiyo na maana;

    2.2 × 10 -5 M = [Ca 2+], [F -] = 0.0013 M

    Angalia:\(\dfrac{2.25×10^{−5}\:M}{0.00133\:M}×100\%=1.69\%\). Thamani hii ni chini ya 5% na inaweza kupuuzwa.

    0.2238 M =\(\ce{[SO4^2- ]}\); [Ag +] = 2.30 × 10 —9 M

    Angalia:\(\dfrac{1.15×10^{−9}}{0.2238}×100\%=5.14×10^{−7}\); hali hiyo imeridhika.

    [OH —] = 2.8 × 10 —3 M; 5.7 × 10 -12 M = [Zn 2+]

    Angalia:\(\dfrac{5.7×10^{−12}}{2.8×10^{−3}}×100\%=2.0×10^{−7}\%\); x ni chini ya 5% ya [OH —] na kwa hiyo, ni kidogo.

    Q15.1.X

    Kwa kuzingatia kwamba hakuna usawa isipokuwa kuvunjwa unahusishwa, uhesabu mkusanyiko wa aina zote za solute katika kila moja ya ufumbuzi wafuatayo wa chumvi unaowasiliana na suluhisho iliyo na ion ya kawaida. Onyesha kwamba mabadiliko katika viwango vya awali vya ions ya kawaida yanaweza kupuuzwa.

    1. TlCl (s) katika 1.250 M HCl
    2. PBi (2 s) katika 0.0355 M CaI 2
    3. Ag 2 CrO 4 (s) katika 0.225 L ya suluhisho iliyo na 0.856 g ya K 2 CrO 4
    4. Cd (OH) 2 (s) katika suluhisho iliyopigwa kwa pH ya 10.995

    Kwa kuzingatia kwamba hakuna usawa isipokuwa kuvunjwa unahusishwa, uhesabu mkusanyiko wa aina zote za solute katika kila moja ya ufumbuzi wafuatayo wa chumvi unaowasiliana na suluhisho iliyo na ion ya kawaida. Onyesha kuwa haifai kupuuza mabadiliko katika viwango vya awali vya ions za kawaida.

    1. TLCl (s) katika 0.025 M TLNo 3
    2. BaF 2 (s) katika 0.0313 M KF
    3. MgC 2 O 4 katika 2.250 L ya suluhisho iliyo na 8.156 g ya Mg (NO 3) 2
    4. Ca (OH) 2 (s) katika suluhisho la unbuffered awali na pH ya 12.700

    S15.1.X

    • [Cl —] = 7.6 × 10 -3 M

    Angalia:\(\dfrac{7.6×10^{−3}}{0.025}×100\%=30\%\)

    Thamani hii ni kubwa mno kuacha x. Kwa hiyo tatua kwa kutumia equation ya quadratic:

    • [Ti +] = 3.1 × 10 —2 M
    • [Cl —] = 6.1 × 10 —3
    • [Ba 2+] = 1.7 × 10 —3 M

    Angalia:\(\dfrac{1.7×10^{−3}}{0.0313}×100\%=5.5\%\)

    Thamani hii ni kubwa mno kushuka x, na equation nzima lazima kutatuliwa.

    • [Ba 2+] = 1.6 × 10 —3 M
    • [F —] = 0.0329 M;
    • Mg (NO 3) 2 = 0.02444 M

    \(\ce{[C2O4^2- ]}=3.5×10^{−3}\)

    Angalia:\(\dfrac{3.5×10^{−3}}{0.02444}×100\%=14\%\)

    Thamani hii ni kubwa kuliko 5%, hivyo equation ya quadratic lazima itumike:

    • \(\ce{[C2O4^2- ]}=3.5×10^{−3}\:M\)
    • [Mg 2+] = 0.0275 M
    • [OH —] = 0.0501 M
    • [Ca 2+] = 3.15 × 10 —3

    Angalia:\(\dfrac{3.15×10^{−3}}{0.050}×100\%=6.28\%\)

    Thamani hii ni kubwa kuliko 5%, hivyo njia halisi zaidi, kama vile makadirio ya mfululizo, lazima itumike.

    • [Ca 2+] = 2.8 × 10 —3 M
    • [OH —] = 0.053 × 10 —2 M

    Q15.1.X

    Eleza kwa nini mabadiliko katika viwango vya ions ya kawaida katika Zoezi yanaweza kupuuzwa.

    Q15.1.X

    Eleza kwa nini mabadiliko katika viwango vya ions ya kawaida katika Zoezi haiwezi kupuuzwa.

    S15.1.X

    Mabadiliko katika mkusanyiko ni makubwa kuliko 5% na hivyo huzidi thamani ya juu ya kupuuza mabadiliko.

    Q15.1.X

    Tumia umumunyifu wa hidroksidi ya alumini, Al (OH) 3, katika suluhisho iliyopigwa kwa pH 11.00.

    Q15.1.X

    Rejea Kiambatisho J kwa bidhaa za umumunyifu kwa chumvi za kalsiamu. Kuamua ni ipi kati ya chumvi za kalsiamu zilizoorodheshwa ni mumunyifu zaidi katika moles kwa lita na ambayo ni mumunyifu zaidi kwa gramu kwa lita.

    S15.1.X

    CaSO 4 2H 2 O ni zaidi mumunyifu Ca chumvi katika mol/L, na pia ni zaidi mumunyifu Ca chumvi katika g/L.

    Q15.1.X

    Misombo ya bariamu nyingi ni sumu sana; hata hivyo, sulfate ya bariamu mara nyingi hutumiwa ndani kama msaada katika uchunguzi wa X-ray wa njia ya chini ya matumbo. Matumizi haya ya BaSO 4 inawezekana kwa sababu ya umumunyifu wake wa chini. Tumia umumunyifu wa molar wa BaSO 4 na wingi wa bariamu uliopo katika 1.00 L ya maji yaliyojaa BaSO 4.

    Q15.1.X

    Viwango vya Huduma za Afya ya Umma kwa maji ya kunywa huweka kiwango cha juu cha 250 mg/L (2.60 × 10 —3 M) au\(\ce{SO4^2-}\) kwa sababu ya hatua yake ya cathartic (ni laxative). Je, maji asilia yanayojaa CaSO 4 (“gyp” maji) kutokana au kupita katika udongo wenye jasi, CaSO 4 •2H 2 O, yanakidhi viwango hivi? Ni nini\(\ce{SO4^2-}\) katika maji kama hayo?

    S15.1.X

    4.9 × 10 —3 M\(\ce{[SO4^2- ]}\) = [Ca 2+]; Kwa kuwa mkusanyiko huu ni wa juu kuliko 2.60 × 10 —3 M, maji ya “gyp” hayafikii viwango.

    Q15.1.X

    Fanya mahesabu yafuatayo:

    1. Tumia [Ag +] katika suluhisho la maji yenye maji ya AGBr.
    2. Je, [Ag +] itakuwa nini wakati KBr ya kutosha imeongezwa ili kufanya [Br —] = 0.050 M?
    3. Nini [Br —] kuwa wakati AgNo 3 ya kutosha imeongezwa kufanya [Ag +] = 0.020 M?

    Bidhaa ya umumunyifu wa CaSO 4 •2H 2 O ni 2.4 × 10 —5. Ni kiasi gani cha chumvi hii itafuta katika 1.0 L ya 0.010 M\(\ce{SO4^2-}\)?

    S15.1.X

    Misa (CaSO 4 •2H 2 O) = 0.34 g/L

    Q15.1.X

    Kutokana kwamba hakuna usawa isipokuwa kuvunjwa ni kushiriki, kuhesabu viwango vya ions katika ufumbuzi ulijaa wa kila moja ya yafuatayo (tazama Jedwali E3 kwa bidhaa umumunyifu).

    1. TLCl
    2. BaF 2
    3. Ag 2 CRO 4
    4. CaC 2 O 4 • H 2 O
    5. anglesite ya madini, PBSO 4

    Q15.1.X

    Kutokana kwamba hakuna usawa isipokuwa kuvunjwa ni kushiriki, kuhesabu viwango vya ions katika ufumbuzi ulijaa wa kila moja ya yafuatayo (tazama Jedwali E3 kwa bidhaa umumunyifu).

    1. AGI
    2. Umri 2 SEO - 4
    3. Mn (OH) 2
    4. Sr (OH) 2 •8H 2 O
    5. brucite madini, Mg (OH) 2

    S15.1.X

    [Ag +] = [I —] = 1.2 × 10 —8 M; [Ag +] = 2.86 × 10 —2 M,\(\ce{[SO4^2- ]}\) = 1.43 × 10 —2 M; [Mn 2+] = 2.2 × 10 —5 M, [OH —] = 4.5 × 10 —5 M; [Sr 2+] = 4.3 × 10 —2 M, [OH —] = 8.6 × 10 —2 M; [Mg 2+] = 1.6 × 10 —4 M, [OH —] = 3.1 × 10 —4 M.

    Q15.1.X

    Viwango vifuatavyo vinapatikana katika mchanganyiko wa ions katika usawa na yabisi kidogo ya mumunyifu. Kutoka kwa viwango vilivyotolewa, hesabu K sp kwa kila moja ya yabisi kidogo ya mumunyifu iliyoonyeshwa:

    1. AGBr: [Ag +] = 5.7 × 10 —7 M, [Br —] = 5.7 × 10 —7 M
    2. CaCO 3: [Ca 2+] = 5.3 × 10 -3 M,\(\ce{[CO3^2- ]}\) = 9.0 × 10 —7 M
    3. PBF 2: [Pb 2+] = 2.1 × 10 —3 M, [F —] = 4.2 × 10 —3 M
    4. Ag 2 CrO 4: [Ag +] = 5.3 × 10 —5 M, 3.2 × 10 —3 M
    5. inF 3: [Katika 3+] = 2.3 × 10 —3 M, [F —] = 7.0 × 10 —3 M

    Q15.1.X

    Viwango vifuatavyo vinapatikana katika mchanganyiko wa ions katika usawa na yabisi kidogo ya mumunyifu. Kutoka kwa viwango vilivyotolewa, hesabu K sp kwa kila moja ya yabisi kidogo ya mumunyifu iliyoonyeshwa:

    1. TLCl: [Tl +] = 1.21 × 10 —2 M, [Cl —] = 1.2 × 10 —2 M
    2. Ce (IO 3) 4: [Ce 4+] = 1.8 × 10 —4 M,\(\ce{[IO3- ]}\) = 2.6 × 10 —13 M
    3. Gd 2 (SO 4) 3: [Gd 3+] = 0.132 M,\(\ce{[SO4^2- ]}\) = 0.198 M
    4. Umri 2 SO 4: [Ag +] = 2.40 × 10 —2 M,\(\ce{[SO4^2- ]}\) = 2.05 × 10 —2 M
    5. BaSO 4: [Ba 2+] = 0.500 M,\(\ce{[SO4^2- ]}\) = 2.16 × 10 —10 M

    S15.1.X

    2.0 × 10 —4; 5.1 × 10 —17; 1.35 × 10 —4; 1.18 × 10 —5; 1.08 × 10 —10

    Q15.1.X

    Ni ipi kati ya misombo yafuatayo inayotokana na suluhisho ambalo lina viwango vinavyoonyeshwa? (Angalia Jedwali E3 kwa maadili K sp.)

    1. KClo 4: [K +] = 0.01 M,\(\ce{[ClO4- ]}\) = 0.01 M
    2. K 2 PtCl 6: [K +] = 0.01 M,\(\ce{[PtCl6^2- ]}\) = 0.01 M
    3. PBi 2: [Pb 2+] = 0.003 M, [I —] = 1.3 × 10 -3 M
    4. Ag 2 S: [Ag +] = 1 × 10 —10 M, [S 2—] = 1 × 10 —13 M

    Q15.1.X

    Ni ipi kati ya misombo yafuatayo inayotokana na suluhisho ambalo lina viwango vinavyoonyeshwa? (Angalia Jedwali E3 kwa maadili K sp.)

    1. CaCO 3: [Ca 2+] = 0.003 M,\(\ce{[CO3^2- ]}\) = 0.003 M
    2. Co (OH) 2: [Co 2+] = 0.01 M, [OH —] = 1 × 10 —7 M
    3. CaPo 4: [Ca 2+] = 0.01 M,\(\ce{[HPO4^2- ]}\) = 2 × 10 —6 M
    4. Pb 3 (PO 4) 2: [Pb 2+] = 0.01 M,\(\ce{[PO4^3- ]}\) = 1 × 10 —13 M

    S15.1.X

    1. CaCO 3 haina precipitate.
    2. Kiwanja hakiingizii.
    3. Kiwanja hakiingizii.
    4. Kiwanja kinazidi.

    Q15.1.X

    Mahesabu ya mkusanyiko wa Tl + wakati TLCl tu huanza precipitate kutoka ufumbuzi yaani 0.0250 M katika Cl —.

    Q15.1.X

    Tumia mkusanyiko wa ion sulfate wakati BaSO 4 inapoanza kuondokana na suluhisho ambalo ni 0.0758 M katika Ba 2+.

    S15.1.X

    1.42 × 10 -9 M

    Q15.1.X

    Mahesabu ya mkusanyiko wa Sr 2+ wakati sRF 2 kuanza precipitate kutoka ufumbuzi yaani 0.0025 M katika F -.

    Q15.1.X

    Tumia mkusanyiko wa\(\ce{PO4^3-}\) wakati Ag 3 PO 4 huanza kuondokana na suluhisho ambalo ni 0.0125 M katika Ag +.

    S15.1.X

    9.2 × 10 -13 M

    Q15.1.X

    Tumia mkusanyiko wa F — inahitajika kuanza mvua ya CaF 2 katika suluhisho ambalo ni 0.010 M katika Ca 2+.

    Q15.1.X

    Tumia mkusanyiko wa Ag + unaotakiwa kuanza mvua ya Ag 2 CO 3 katika suluhisho ambalo ni 2.50 × 10 —6 M ndani\(\ce{CO3^2-}\).

    S15.1.X

    [Umri +] = 1.8 × 10 —3 M

    Q15.1.X

    Nini [Ag +] inahitajika ili kupunguza\(\ce{[CO3^2- ]}\) hadi 8.2 × 10 —4 M kwa mvua ya Ag 2 CO 3?

    Q15.1.X

    Nini [F —] inahitajika ili kupunguza [Ca 2+] hadi 1.0 × 10 —4 M kwa mvua ya CaF 2?

    S15.1.X

    6.2 × 10 —4

    Q15.1.X

    Kiasi cha 0.800 L ya 2 × 10 —4 - M Ba (NO 3) 2 ufumbuzi ni aliongeza kwa 0.200 L ya 5 × 10 —4 M Li 2 SO 4. Je, BaSO 4 huzuia? Eleza jibu lako.

    Q15.1.X

    Fanya mahesabu haya kwa carbonate ya nickel (II).

    1. Kwa kiasi gani cha maji lazima iwe na usahihi unao na NiCO 3 uoshwe ili kufuta 0.100 g ya kiwanja hiki? Fikiria kwamba maji ya safisha yanajaa NiCo 3 (K sp = 1.36 × 10 —7).
    2. Ikiwa NiCo 3 walikuwa mchafuzi katika sampuli ya CoCO 3 (K sp = 1.0 × 10 —12), ni umati gani wa CoCO 3 ungepotea? Kumbuka kwamba wote NiCO 3 na CoCO 3 kufuta katika suluhisho moja.

    S15.1.X

    2.28 L; 7.3 × 10 —7 g

    Q15.1.X

    Viwango vya chuma vikubwa zaidi ya 5.4 × 10 —6 M katika maji vinavyotumika kwa madhumuni ya kufulia vinaweza kusababisha uchafu. Nini [OH —] inahitajika ili kupunguza [Fe 2+] kwa kiwango hiki kwa mvua ya Fe (OH) 2?

    Q15.1.X

    Suluhisho ni 0.010 M katika Cu 2+ na Cd 2+. Ni asilimia gani ya Cd 2+ inabakia katika suluhisho wakati 99.9% ya Cu 2+ imesababishwa kama CU kwa kuongeza sulfidi?

    S15.1.X

    100% ya hiyo ni kufutwa

    Q15.1.X

    Suluhisho ni 0.15 M katika Pb 2+ na Ag +. Kama Cl — ni aliongeza kwa ufumbuzi huu, ni nini [Ag +] wakati PBCl 2 huanza precipitate?

    Q15.1.X

    Ni reagent gani inayoweza kutumika kutenganisha ions katika kila mchanganyiko wafuatayo, ambayo ni 0.1 M kwa heshima ya kila ion? Katika hali nyingine inaweza kuwa muhimu kudhibiti pH. (Kidokezo: Fikiria maadili ya K sp yaliyotolewa katika Kiambatisho J.)

    1. \(\ce{Hg2^2+}\)na Cu 2+
    2. \(\ce{SO4^2-}\)na Cl —
    3. Hg 2+ na Co 2+
    4. Zn 2+ na Sr 2+
    5. Ba 2+ na Mg 2+
    6. \(\ce{CO3^2-}\)na OH —

    S15.1.X

    1. \(\ce{Hg2^2+}\)na Cu 2+: Kuongeza\(\ce{SO4^2-}\).
    2. \(\ce{SO4^2-}\)na Cl -: Kuongeza Ba 2+.
    3. Hg 2+ na Co 2+: Ongeza S 2—.
    4. Zn 2+ Sr 2+: Kuongeza OH — mpaka [OH -] = 0.050 M.
    5. Ba 2+ na Mg 2+: Kuongeza\(\ce{SO4^2-}\).
    6. \(\ce{CO3^2-}\)na OH -: Kuongeza Ba 2+.

    Q15.1.X

    Suluhisho lina 1.0 × 10 —5 mol ya KBR na 0.10 mol ya KCl kwa lita. AgNo 3 ni hatua kwa hatua aliongeza kwa suluhisho hili. Ni aina ya kwanza, AGBr imara au agCl imara?

    Q15.1.X

    Suluhisho lina 1.0 × 10 —2 mol ya KI na 0.10 mol ya kCl kwa lita. AgNo 3 ni hatua kwa hatua aliongeza kwa suluhisho hili. Ni aina ya kwanza, AGI imara au agCl imara?

    S15.1.X

    AGI itakuwa precipitate kwanza.

    Q15.1.X

    Ions ya kalsiamu katika seramu ya damu ya binadamu ni muhimu kwa kuchanganya. Oxalate ya potassium, K 2 C 2 O 4, hutumiwa kama anticoagulant wakati sampuli ya damu inapokanzwa kwa vipimo vya maabara kwa sababu huondoa calcium kama precipitate ya CaC 2 O 4 • H 2 O. ni muhimu kuondoa wote lakini 1.0% ya Ca 2+ katika serum ili kuzuia kuchanganya. Ikiwa seramu ya kawaida ya damu yenye pH iliyozuiliwa ya 7.40 ina 9.5 mg ya Ca 2+ kwa 100 ml ya serum, ni kiasi gani cha K 2 C 2 O 4 kinachohitajika ili kuzuia mgando wa sampuli ya damu ya 10 ml ambayo ni serum 55% kwa kiasi? (Wote kiasi ni sahihi kwa takwimu mbili muhimu. Kumbuka kwamba kiasi cha seramu katika sampuli ya damu 10-ml ni 5.5 ml. Fikiria kwamba thamani ya K sp kwa CaC 2 O 4 katika serum ni sawa na katika maji.)

    Q15.1.X

    Kuhusu 50% ya calculi ya mkojo (mawe ya figo) yanajumuisha phosphate ya kalsiamu, Ca 3 (PO 4) 2. Maudhui ya kawaida ya kalsiamu ya kati yaliyopunguzwa katika mkojo ni 0.10 g ya Ca 2+ kwa siku. Kiwango cha kawaida cha kati cha mkojo kilichopitishwa kinaweza kuchukuliwa kama 1.4 L kwa siku. Je, ni mkusanyiko wa juu wa ioni ya phosphate ambayo mkojo unaweza kuwa na kabla ya calculus kuanza kuunda?

    S15.1.X

    4 × 10 -9 M

    Q15.1.X

    PH ya mkojo wa kawaida ni 6.30, na mkusanyiko wa phosphate jumla (\(\ce{[PO4^3- ]}\)\(\ce{[HPO4^2- ]}\)\(\ce{[H2PO4- ]}\)+ + + [H 3 PO 4]) ni 0.020 M. Je, ni mkusanyiko wa chini wa Ca 2+muhimu ili kushawishi malezi ya mawe ya figo? (Angalia Zoezi kwa maelezo ya ziada.)

    Q15.1.X

    Metali ya magnesiamu (sehemu ya aloi zinazotumiwa katika ndege na wakala wa kupunguza kutumika katika uzalishaji wa uranium, titani, na metali nyingine za kazi) hutengwa na maji ya bahari kwa mlolongo wa athari zifuatazo:

    \(\ce{Mg^2+}(aq)+\ce{Ca(OH)2}(aq)⟶\ce{Mg(OH)2}(s)+\ce{Ca^2+}(aq)\)

    \(\ce{Mg(OH)2}(s)+\ce{2HCl}(aq)⟶\ce{MgCl2}(s)+\ce{2H2O}(l)\)

    \(\ce{MgCl2}(l)\xrightarrow{\ce{electrolysis}}\ce{Mg}(s)+\ce{Cl2}(g)\)

    Maji ya bahari yana wiani wa 1.026 g/cm 3 na ina sehemu 1272 kwa milioni ya magnesiamu kama Mg 2+ (aq) kwa wingi. Masi gani, kwa kilo, ya Ca (OH) 2 inahitajika kuziba 99.9% ya magnesiamu katika 1.00 × 10 3 L ya maji ya bahari?

    S15.1.X

    3.99 kilo

    Q15.1.X

    Sulfidi hidrojeni hupigwa ndani ya suluhisho ambalo ni 0.10 M katika Pb 2+ na Fe 2+ na 0.30 M katika HCl. Baada ya ufumbuzi umefika kwa usawa imejaa H 2 S ([H 2 S] = 0.10 M). Ni viwango gani vya Pb 2+ na Fe 2+ vinavyobaki katika suluhisho? Kwa ufumbuzi uliojaa wa H 2 S tunaweza kutumia usawa:

    \(\ce{H2S}(aq)+\ce{2H2O}(l)⇌\ce{2H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K=1.0×10^{−26}\)

    (Kidokezo:\(\ce{[H3O+]}\) mabadiliko kama sulfidi chuma precipitate.)

    Q15.1.X

    Fanya mahesabu yafuatayo yanayohusisha viwango vya ioni za iodate:

    1. Mkusanyiko wa iodate ion ya ufumbuzi uliojaa wa La (IO 3) 3 ulipatikana kuwa 3.1 × 10 —3 mol/L.
    2. Pata mkusanyiko wa ions iodate katika suluhisho iliyojaa ya Cu (IO 3) 2 (K sp = 7.4 × 10 —8).

    S15.1.X

    3.1 × 10 —11; [Cu 2+] = 2.6 × 10 —3;\(\ce{[IO3- ]}\) = 5.3 × 10 —3

    Q15.1.X

    Tumia umumunyifu wa molar wa AgBr katika 0.035 M NaBr (K sp = 5 × 10 —13).

    Q15.1.X

    Ni gramu ngapi za Pb (OH) 2 zitapasuka katika 500 ml ya ufumbuzi wa 0.050- M PBCl 2 (K sp = 1.2 × 10 —15)?

    S15.1.X

    1.8 × 10 —5 g Pb (OH) 2

    Q15.1.X

    Matumizi simulation kutoka Link mapema kwa Learning kukamilisha zoezi zifuatazo:. Kutumia 0.01 g CaF 2, fanya maadili ya K sp yaliyopatikana katika suluhisho la 0.2- M la kila chumvi. Jadili kwa nini maadili hubadilika unapobadilisha chumvi za mumunyifu.

    Q15.1.X

    Ni gramu ngapi za Maziwa ya Magnesia, Mg (OH) 2 (s) (58.3 g/mol), ingekuwa mumunyifu katika 200 ml ya maji. K sp = 7.1 × 10 —12. Jumuisha majibu ya ionic na usemi wa K sp katika jibu lako. (K w = 1 × 10 —14 = [H 3 O +] [OH -])

    S15.1.X

    \[\ce{Mg(OH)2}(s)⇌\ce{Mg^2+}+\ce{2OH-} \]

    \[K_\ce{sp}=\ce{[Mg^2+][OH- ]^2}\]

    \[1.14 × 10−3 g Mg(OH)2\]

    Q15.1.X

    Chumvi mbili za nadharia, LM 2 na LQ, zina umumunyifu sawa wa molar katika H 2 O. ikiwa K sp kwa LM 2 ni 3.20 × 10 —5, ni thamani gani ya K sp kwa LQ?

    Q15.1.X

    Ni ipi kati ya carbonates zifuatazo zitaunda kwanza? Ni ipi kati ya yafuatayo itaunda mwisho? Eleza.

    1. \(\ce{MgCO3} \hspace{20px} K_\ce{sp}=3.5×10^{−8}\)
    2. \(\ce{CaCO3} \hspace{20px} K_\ce{sp}=4.2×10^{−7}\)
    3. \(\ce{SrCO3} \hspace{20px} K_\ce{sp}=3.9×10^{−9}\)
    4. \(\ce{BaCO3} \hspace{20px} K_\ce{sp}=4.4×10^{−5}\)
    5. \(\ce{MnCO3} \hspace{20px} K_\ce{sp}=5.1×10^{−9}\)

    S15.1.X

    SRCo 3 fomu ya kwanza, kwa kuwa ina ndogo K sp thamani ni angalau mumunyifu. BaCO 3 itakuwa ya mwisho ya kuharakisha, ina thamani kubwa ya K sp.

    Q15.1.X

    Ni gramu ngapi za Zn (CN) 2 (s) (117.44 g/mol) zitakuwa mumunyifu katika 100 ml ya H 2 O? Ni pamoja na majibu uwiano na kujieleza kwa K sp katika jibu lako. Thamani ya K sp kwa Zn (CN) 2 (s) ni 3.0 × 10 —16.

    15.2: Lewis Acids na Msingi

    Q15.2.X

    Chini ya hali gani, ikiwa ipo, je, sampuli ya AgCl imara hupasuka kabisa katika maji safi?

    S15.2.X

    wakati kiasi cha imara ni ndogo sana kwamba ufumbuzi ulijaa si zinazozalishwa

    Q15.2.X

    Eleza kwa nini kuongeza NH 3 au HNO 3 kwa suluhisho iliyojaa ya Ag 2 CO 3 katika kuwasiliana na imara Ag 2 CO 3 huongeza umumunyifu wa imara.

    Q15.2.X

    Tumia mkusanyiko wa ioni ya cadmium, [Cd 2+], katika suluhisho lililoandaliwa kwa kuchanganya 0.100 L ya 0.100 M Cd (NO 3) 2 na 1.150 L ya 0.100 NH 3 (aq).

    S15.2.X

    2.35 × 10 —4 M

    Q15.2.X

    Eleza kwa nini kuongeza NH 3 au HNO 3 kwa suluhisho iliyojaa ya Cu (OH) 2 katika kuwasiliana na imara Cu (OH) 2 huongeza umumunyifu wa imara.

    S15.2.X

    Wakati mwingine usawa kwa ions tata huelezwa kwa suala la kutofautiana mara kwa mara, K d. Kwa ioni tata mmenyuko\(\ce{AlF6^3-}\) wa dissociation ni:

    \[\ce{AlF6^3- ⇌ Al^3+ + 6F-}\) and \(K_\ce{d}=\ce{\dfrac{[Al^3+][F- ]^6}{[AlF6^3- ]}}=2×10^{−24}\]

    Q15.2.X

    Tumia thamani ya mara kwa mara ya malezi, K f, kwa\(\ce{AlF6^3-}\).

    S15.2.X

    5 × 10 23

    Q15.2.X

    Kutumia thamani ya mara kwa mara ya malezi kwa ion tata\(\ce{Co(NH3)6^2+}\), kuhesabu mara kwa mara dissociation.

    Q15.2.X

    Kutumia mara kwa mara dissociation, K d = 7.8 × 10 —18, mahesabu ya viwango vya usawa wa Cd 2+ na CN — katika ufumbuzi wa 0.250- M ya\(\ce{Cd(CN)4^2-}\).

    S15.2.X

    Jedwali hili lina nguzo mbili kuu na safu tatu. Mstari wa kwanza wa safu ya kwanza haina kichwa na kisha ina zifuatazo katika safu ya kwanza: Mkusanyiko wa awali (M) na Msawazo (M). Safu ya pili ina kichwa, “[C d (C N) subscript 4 kwa ishara ya pili ya nguvu superscript hasi] [C N superscript ishara hasi] [C d kwa nguvu ya pili superscript ishara chanya].” Chini ya safu ya pili ni kikundi cha safu mbili na nguzo tatu. Safu ya kwanza ina yafuatayo: 0.250 na 0.250 bala x. safu ya pili ina yafuatayo: 0 na 4 x safu ya tatu ina yafuatayo: 0 na x.

    [Cd 2+] = 9.5 × 10 —5 M; [CN —] = 3.8 × 10 —4 M

    Q15.2.X

    Kutumia mara kwa mara ya kujitenga, K d = 3.4 × 10 —15, mahesabu ya viwango vya usawa wa Zn 2+ na OH — katika suluhisho la 0.0465- M la\(\ce{Zn(OH)4^2-}\).

    Q15.2.X

    Using the dissociation constant, Kd = 2.2 × 10–34, calculate the equilibrium concentrations of Co3+ and NH3 in a 0.500-M solution of \(\ce{Co(NH3)6^3+}\).

    S15.2.X

    [Co3+] = 3.0 × 10–6 M; [NH3] = 1.8 × 10–5 M

    Q15.2.X

    Using the dissociation constant, Kd = 1 × 10–44, calculate the equilibrium concentrations of Fe3+ and CN– in a 0.333 M solution of \(\ce{Fe(CN)6^3-}\).

    Q15.2.X

    Calculate the mass of potassium cyanide ion that must be added to 100 mL of solution to dissolve 2.0 × 10–2 mol of silver cyanide, AgCN.

    S15.2.X

    1.3 g

    Q15.2.X

    Calculate the minimum concentration of ammonia needed in 1.0 L of solution to dissolve 3.0 × 10–3 mol of silver bromide.

    Q15.2.X

    A roll of 35-mm black and white photographic film contains about 0.27 g of unexposed AgBr before developing. What mass of Na2S2O3•5H2O (sodium thiosulfate pentahydrate or hypo) in 1.0 L of developer is required to dissolve the AgBr as \(\ce{Ag(S2O3)2^3-}\) (Kf = 4.7 × 1013)?

    S15.2.X

    0.80 g

    Q15.2.X

    We have seen an introductory definition of an acid: An acid is a compound that reacts with water and increases the amount of hydronium ion present. In the chapter on acids and bases, we saw two more definitions of acids: a compound that donates a proton (a hydrogen ion, H+) to another compound is called a Brønsted-Lowry acid, and a Lewis acid is any species that can accept a pair of electrons. Explain why the introductory definition is a macroscopic definition, while the Brønsted-Lowry definition and the Lewis definition are microscopic definitions.

    Q15.2.X

    Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each:

    1. \(\ce{CO2 + OH- ⟶ HCO3-}\)
    2. \(\ce{B(OH)3 + OH- ⟶ B(OH)4-}\)
    3. \(\ce{I- + I2 ⟶ I3-}\)
    4. \(\ce{AlCl3 + Cl- ⟶ AlCl4-}\) (use Al-Cl single bonds)
    5. \(\ce{O^2- + SO3 ⟶ SO4^2-}\)

    S15.2.X

    (a)

    This figure shows a chemical reaction modeled with structural formulas. On the left side is a structure with a central C atom. O atoms, each with two unshared electron pairs, are double bonded to the left and right sides of the C atom. Following a plus sign is another structure in brackets which has an O atom with three unshared electron dot pairs single bonded to an H atom on the right. Outside the brackets is superscript negative sign. Following a right pointing arrow is a structure in brackets that has a central C atom to which 3 O atoms are bonded. Above and slightly to the right, one of the O atoms is connected with a double bond. This O atom has two unshared electron pairs. The second O atom is single bonded below and slightly to the right. This O atom has three unshared electron pairs. The third O atom is bonded to the left of the C atom. This O atom has two unshared electron pairs and an H atom single bonded to its left. Outside the brackets to the right is a superscript negative symbol. ;

    (b)

    Takwimu hii inaonyesha mmenyuko wa kemikali unaoelekezwa na formula za kimuundo. Upande wa kushoto ni muundo ambao una atomi ya kati ya B ambayo atomi za O 3 zinaunganishwa. Atomi O juu na chini ya haki kidogo ya atomu B kila mmoja zina moja ya atomu H iliyofungwa upande wa kulia. Atomu ya tatu O ni moja iliyofungwa upande wa kushoto wa atomu B. Atomi hii O ina atomu ya H moja iliyounganishwa upande wake wa kushoto. Atomi zote O katika muundo huu zina jozi mbili za elektroni zisizoshirikishwa. Kufuatia ishara pamoja ni muundo mwingine ambao una O atomu moja iliyounganishwa na atomu H upande wake wa kulia. Atomu O ina jozi tatu za elektroni zisizoshirikiwa. Mfumo unaonekana katika mabano na ishara mbaya ya superscript. Kufuatia kulia akizungumzia mshale ni muundo katika mabano ina kati B atomi ambayo 4 O atomi ni bonded. Atomi O juu, chini, na haki ya atomi B kila mmoja ana H atomi moja iliyofungwa na haki. Atomu ya tatu O ni moja iliyofungwa upande wa kushoto wa atomu B. Atomi hii O ina atomu ya H moja iliyounganishwa upande wake wa kushoto. Atomi zote O katika muundo huu zina jozi mbili za elektroni zisizoshirikishwa. Nje ya mabano kwa haki ni ishara hasi ya superscript.;

    (c)

    Takwimu hii inaonyesha mmenyuko wa kemikali kwa kutumia formula za miundo. Kwenye upande wa kushoto, atomi mbili za I, kila mmoja na jozi tatu za elektroni zisizo na ushirikiano, zimeunganishwa na dhamana moja. Kufuatia ishara ya pamoja ni muundo mwingine ambao una atomi I na jozi nne za dots za elektroni na ishara hasi ya superscript. Kufuatia kulia akizungumzia mshale ni muundo katika mabano ambayo ina tatu mimi atomi kushikamana katika mstari na vifungo moja. mbili mwisho mimi atomi na tatu unshared elektroni dot jozi na mimi atomi katika kituo ina mbili unshared elektroni jozi. Nje ya mabano ni ishara mbaya ya superscript.;

    (d)

    Takwimu hii inaonyesha mmenyuko wa kemikali kwa kutumia formula za miundo. Kwenye upande wa kushoto, atomi ya L imewekwa katikati ya muundo na atomi tatu za Cl ni moja iliyofungwa hapo juu, leftt, na chini. Kila atomi ya C l ina jozi tatu za dots za elektroni. Kufuatia ishara ya pamoja ni muundo mwingine ambao una atomi F umezungukwa na jozi nne za elektroni nukta na alama hasi ya superscript. Kufuatia mshale unaoelekeza kulia ni muundo katika mabano ambayo ina atomi ya kati ya L ambayo atomi 4 C l zinaunganishwa na vifungo moja hapo juu, chini, upande wa kushoto, na kulia. Kila atomi ya C l katika muundo huu ina jozi tatu za dots za elektroni. Nje ya mabano ni ishara mbaya ya superscript.;

    (e)

    Takwimu hii inaonyesha mmenyuko wa kemikali kwa kutumia formula za miundo. Upande wa kushoto ni muundo ambao una atomi ya S katikati. O atomi ni moja bonded juu na chini. Atomi hizi O zina jozi tatu za nukta za elektroni kila mmoja. Kwa haki ya atomu ya S ni atomi ya O iliyofungwa mara mbili ambayo ina jozi mbili za nukta za elektroni. Kufuatia ishara ya pamoja ni atomi O ambayo imezungukwa na jozi nne za nukta za elektroni na ina superscript 2 hasi. Kufuatia mshale unaoelekeza kulia ni muundo katika mabano ambayo ina atomi ya kati ya S ambayo atomi 4 O zinaunganishwa na vifungo moja hapo juu, chini, upande wa kushoto, na kulia. Kila moja ya atomi O ina jozi tatu za nukta za elektroni. Nje ya mabano ni superscript 2 hasi.

    Q15.2.X

    Andika miundo Lewis ya reactants na bidhaa ya kila moja ya milinganyo zifuatazo, na kutambua asidi Lewis na msingi Lewis katika kila:

    1. \(\ce{CS2 + SH- ⟶ HCS3-}\)
    2. \(\ce{BF3 + F- ⟶ BF4-}\)
    3. \(\ce{I- + SnI2 ⟶ SnI3-}\)
    4. \(\ce{Al(OH)3 + OH- ⟶ Al(OH)4-}\)
    5. \(\ce{F- + SO3 ⟶ SFO3-}\)

    Q15.2.X

    Using Lewis structures, write balanced equations for the following reactions:

    1. \(\ce{HCl}(g)+\ce{PH3}(g)⟶\)
    2. \(\ce{H3O+ + CH3- ⟶}\)
    3. \(\ce{CaO + SO3 ⟶}\)
    4. \(\ce{NH4+ + C2H5O- ⟶}\)

    S15.2.X

    (a)

    This figure represents a chemical reaction in two rows. The top row shows the reaction using chemical formulas. The second row uses structural formulas to represent the reaction. The first row contains the equation H C l ( g ) plus P H subscript 3 ( g ) right pointing arrow left bracket P H subscript 4 right bracket superscript plus plus left bracket C l with 4 pairs of electron dots right bracket superscript negative sign. The second row begins on the left with H left bracket C l with four unshared electron pairs right bracket plus a structure in brackets with a central P atom with H atoms single bonded at the left, above, and to the right. A single unshared electron pair is on the central P atom. Outside the brackets to the right is a superscript plus sign. Following a right pointing arrow is a structure in brackets with a central P atom with H atoms single bonded at the left, above, below, and to the right. Outside the brackets is a superscript plus sign. This structure is followed by a plus and a C l atom in brackets with four unshared electron pairs and a superscript negative sign. ;

    \(\ce{H3O+ + CH3- ⟶ CH4 + H2O}\)

    This figure represents a chemical reaction using structural formulas. A structure is shown in brackets on the left which is composed of a central O atom with one unshared electron pair and three single bonded H atoms to the left, right, and above the atom. Outside the brackets to the right is a superscript plus sign. Following a plus sign, is another structure in brackets composed of a central C atom with one unshared electron pair and three single bonded H atoms to the left, right, and above the atom. Outside the brackets to the right is a superscript negative sign. Following a right pointing arrow is a structure with a central C atom with H atoms single bonded above, below, left and right. Following a plus sign is a structure with a central O atom with two unshared electron pairs and two H atoms connected with single bonds. ;

    \(\ce{CaO + SO3 ⟶ CaSO4}\)

    This figure represents a chemical reaction using structural formulas. On the left, C a superscript 2 plus is just left of bracket O with four unshared electron pairs right bracket superscript 2 negative plus a structure with a central S atom to which two O atoms are single bonded at the left and right, and a single O atom is double bonded above. The two single bonded O atoms each have three unshared electron pairs and the double bonded O atom has two unshared electron pairs. Following a right pointing arrow is C a superscript 2 plus just left of a structure in brackets with a central S atom which has 4 O atoms single bonded at the left, above, below, and to the right. Each of the O atoms has three unshared electron pairs. Outside the brackets to the right is a superscript two negative. ;

    \(\ce{NH4+ + C2H5O- ⟶ C2H5OH + NH3}\)

    This figure represents a chemical reaction using structural formulas. A structure is shown in brackets on the left which is composed of a central N atom with four single bonded H atoms to the left, right, above, and below the atom. Outside the brackets to the right is a superscript plus sign. Following a plus sign, is another structure in brackets composed of a C atom with three single bonded H atoms above, below, and to the left. A second C atom is single bonded to the right. This C atom has H atoms single bonded above and below. To the right of the second C atom, an O atom is single bonded. This O atom has three unshared electron pairs. Outside the brackets to the right is a subperscript negative. Following a right pointing arrow is a structure composed of a C atom with three single bonded H atoms above, below, and to the left. A second C atom is single bonded to the right. This C atom has H atoms single bonded above and below. To the right of the second C atom, an O atom is single bonded. This O atom has two unshared electron pairs and an H atom single bonded to its right.

    Q15.2.X

    Tumia\(\ce{[HgCl4^2- ]}\) katika suluhisho iliyoandaliwa kwa kuongeza 0.0200 mol ya NaCl hadi 0.250 L ya ufumbuzi wa 0.100- M HgCl 2.

    Q15.2.X

    Katika titration ya ioni ya cyanide, 28.72 ml ya 0.1000 M AgNo 3 huongezwa kabla ya mvua kuanza. [Majibu ya Ag + na CN — inakwenda kukamilika, kuzalisha\(\ce{Ag(CN)2-}\) tata.] Upepo wa AGCN imara unafanyika wakati ziada Ag + ni aliongeza kwa ufumbuzi, juu ya kiasi kinachohitajika kukamilisha malezi ya\(\ce{Ag(CN)2-}\). Ni gramu ngapi za NaCN zilikuwa katika sampuli ya awali?

    S15.2.X

    0.0281 g

    Q15.2.X

    Je, ni viwango vya Ag +, CN —, na\(\ce{Ag(CN)2-}\) katika suluhisho iliyojaa ya AgCn?

    Q15.2.X

    Katika ufumbuzi wa maji machafu HF hufanya kama asidi dhaifu. Hata hivyo, kioevu safi HF (kiwango cha kuchemsha = 19.5 °C) ni asidi kali. Katika HF ya kioevu, HNO 3 hufanya kama msingi na inakubali protoni. Asidi ya HF kiowevu inaweza kuongezeka kwa kuongeza moja ya fluoridi kadhaa isokaboni ambazo ni asidi Lewis na kukubali F — ion (kwa mfano, BF 3 au SBF 5). Andika usawa wa kemikali kwa mmenyuko wa HNO safi 3 na HF safi na ya HF safi na BF 3.

    S15.2.X

    \(\ce{HNO3}(l)+\ce{HF}(l)⟶\ce{H2NO3+}+\ce{F-}\);\(\ce{HF}(l)+\ce{BF3}(g)⟶\ce{H+}+\ce{BF4}\)

    Q15.2.X

    Asidi amino rahisi ni glycine, H 2 NCH 2 CO 2 H. kipengele cha kawaida cha asidi amino ni kwamba zina makundi ya kazi: kundi la amine, -NH 2, na kundi la asidi ya kaboksili, -CO 2 H. asidi amino inaweza kufanya kazi kama asidi au msingi. Kwa glycine, nguvu ya asidi ya kundi la carboxyl ni sawa na ile ya asidi ya asidi, CH 3 CO 2 H, na nguvu ya msingi ya kundi la amino ni kubwa zaidi kuliko ile ya amonia, NH 3.

    Q15.2.X

    Andika miundo ya Lewis ya ions ambayo huunda wakati glycine inapasuka katika 1 M HCl na 1 M KOH.

    Q15.2.X

    Andika muundo wa Lewis wa glycine wakati asidi hii ya amino inapasuka katika maji. (Kidokezo: Fikiria nguvu za msingi za -NH 2 na\(\ce{−CO2-}\) vikundi.)

    Q15.2.X

    Asidi ya boroni, H 3 BO 3, si asidi ya Brønsted-Lowry bali asidi ya Lewis.

    1. Andika equation kwa majibu yake na maji.
    2. Kutabiri sura ya anion hivyo sumu.
    3. Je, ni hybridization juu ya boron sambamba na sura una alitabiri?

    S15.2.X

    \(\ce{H3BO3 + H2O ⟶ H4BO4- + H+}\); Maumbo ya umeme na Masi ni sawa-wote tetrahedral. Muundo wa tetrahedral ni sawa na sp 3 hybridization.

    15.3: Multiple Mlinganyo

    Q15.3.1

    Suluhisho lililojaa la electrolyte kidogo ya mumunyifu katika kuwasiliana na baadhi ya electrolyte imara inasemekana kuwa mfumo katika usawa. Eleza. Kwa nini mfumo kama huo unaitwa usawa usio na kawaida?

    Q15.3.2

    Tumia mkusanyiko wa usawa wa Ni 2+ katika suluhisho la 1.0- M [Ni (NH 3) 6] (NO 3) 2.

    S15.3.2

    0.014 M

    Q15.3.3

    Tumia mkusanyiko wa usawa wa Zn 2+ katika ufumbuzi wa 0.30- M wa\(\ce{Zn(CN)4^2-}\).

    Q15.3.4

    Tumia mkusanyiko wa usawa wa Cu 2+ katika suluhisho awali na 0.050 M Cu 2+ na 1.00 M NH 3.

    S15.3.2

    1.0 × 10 —13 M

    Q15.3.5

    Tumia msawazo mkusanyiko wa Zn 2+ katika suluhisho awali na 0.150 M Zn 2+ na 2.50 M CN -.

    Q15.3.6

    Tumia mkusanyiko wa usawa wa Fe 3+ wakati 0.0888 mole ya K 3 [Fe (CN) 6] imeongezwa kwenye suluhisho na 0.0.00010 M CN -.

    S15.3.2

    9 × 10 -22 M

    Q15.3.7

    Tumia mkusanyiko wa usawa wa Co 2+ wakati 0.100 mole ya [Co (NH 3) 6] (NO 3) 2 imeongezwa kwenye suluhisho na 0.025 M NH 3. Fikiria kiasi ni 1.00 L.

    Q15.3.8

    Mara kwa mara ya usawa kwa mmenyuko\(\ce{Hg^2+}(aq)+\ce{2Cl-}(aq)⇌\ce{HgCl2}(aq)\) ni 1.6 × 10 13. HgCl 2 ni electrolyte kali au electrolyte dhaifu? Je, ni viwango vya Hg 2+ na Cl — katika ufumbuzi wa 0.015- M wa HgCl 2?

    S15.3.2

    6.2 × 10 —6 M = [Hg 2+]; 1.2 × 10 —5 M = [Cl —]; Dutu hii ni electrolyte dhaifu kwa sababu kidogo sana ya awali 0.015 M HgCl 2 kufutwa.

    Q15.3.9

    Tumia umumunyifu wa molar wa Sn (OH) 2 katika suluhisho la buffer iliyo na viwango sawa vya NH 3 na\(\ce{NH4+}\).

    Q15.3.X

    Tumia umumunyifu wa molar wa Al (OH) 3 katika suluhisho la buffer na 0.100 M NH 3 na 0.400 M\(\ce{NH4+}\).

    S15.3.2

    [OH -] = 4.5 × 10 -5; [Al 3+] = 2.1 × 10 —20 (umumunyifu wa molar)

    Q15.3.10 Badilisha sehemu

    Je, ni umumunyifu wa molar wa CaF 2 katika ufumbuzi wa 0.100- M wa HF? K a kwa HF = 7.2 × 10 —4.

    Q15.3.X

    Je, ni umumunyifu wa molar wa BaSO 4 katika ufumbuzi wa 0.250- M wa NaHSO 4? K a kwa\(\ce{HSO4-}\) = 1.2 × 10–2.

    S15.3.2

    • \(\ce{[SO4^2- ]}=0.049\:M\)
    • [Ba2+] = 2.2 × 10–9 (molar solubility)

    Q15.3.X

    What is the molar solubility of Tl(OH)3 in a 0.10-M solution of NH3?

    Q15.3.X

    What is the molar solubility of Pb(OH)2 in a 0.138-M solution of CH3NH2?

    S15.3.2

    • [OH–] = 7.6 × 10−3 M
    • [Pb2+] = 4.8 × 10–12 (molar solubility)

    Q15.3.X

    A solution of 0.075 M CoBr2 is saturated with H2S ([H2S] = 0.10 M). What is the minimum pH at which CoS begins to precipitate?

    \(\ce{CoS}(s)⇌\ce{Co^2+}(aq)+\ce{S^2-}(aq) \hspace{20px} K_\ce{sp}=4.5×10^{−27}\)

    \(\ce{H2S}(aq)+\ce{2H2O}(l)⇌\ce{2H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K=1.0×10^{−26}\)

    A 0.125-M solution of Mn(NO3)2 is saturated with H2S ([H2S] = 0.10 M). At what pH does MnS begin to precipitate?

    \(\ce{MnS}(s)⇌\ce{Mn^2+}(aq)+\ce{S^2-}(aq) \hspace{20px} K_\ce{sp}=4.3×10^{−22}\)

    \(\ce{H2S}(aq)+\ce{2H2O}(l)⇌\ce{2H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K=1.0×10^{−26}\)

    S15.3.2

    3.27

    Q15.3.X

    Calculate the molar solubility of BaF2 in a buffer solution containing 0.20 M HF and 0.20 M NaF.

    Q15.3.X

    Calculate the molar solubility of CdCO3 in a buffer solution containing 0.115 M Na2CO3 and 0.120 M NaHCO3

    S15.3.2

    • \(\ce{[CO3^2- ]}=0.115\:M\)
    • [Cd2+] = 3 × 10−12 M

    Q15.3.X

    To a 0.10-M solution of Pb(NO3)2 is added enough HF(g) to make [HF] = 0.10 M.

    1. Does PbF2 precipitate from this solution? Show the calculations that support your conclusion.
    2. What is the minimum pH at which PbF2 precipitates?

    Calculate the concentration of Cd2+ resulting from the dissolution of CdCO3 in a solution that is 0.010 M in H2CO3.

    S15.3.2

    1 × 10−5 M

    Q15.3.X

    Both AgCl and AgI dissolve in NH3.

    1. What mass of AgI dissolves in 1.0 L of 1.0 M NH3?
    2. What mass of AgCl dissolves in 1.0 L of 1.0 M NH3?

    Calculate the volume of 1.50 M CH3CO2H required to dissolve a precipitate composed of 350 mg each of CaCO3, SrCO3, and BaCO3.

    S15.3.2

    0.0102 L (10.2 mL)

    Q15.3.X

    Even though Ca(OH)2 is an inexpensive base, its limited solubility restricts its use. What is the pH of a saturated solution of Ca(OH)2?

    Q15.3.X

    What mass of NaCN must be added to 1 L of 0.010 M Mg(NO3)2 in order to produce the first trace of Mg(OH)2?

    S15.3.2

    5 × 10−3 g

    Q15.3.X

    Magnesium hydroxide and magnesium citrate function as mild laxatives when they reach the small intestine. Why do magnesium hydroxide and magnesium citrate, two very different substances, have the same effect in your small intestine. (Hint: The contents of the small intestine are basic.)

    Q15.3.X

    The following question is taken from a Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service.

    Solve the following problem:

    \(\ce{MgF2}(s)⇌\ce{Mg^2+}(aq)+\ce{2F-}(aq)\)

    In a saturated solution of MgF2 at 18 °C, the concentration of Mg2+ is 1.21 × 10–3 M. The equilibrium is represented by the preceding equation.

    1. Write the expression for the solubility-product constant, Ksp, and calculate its value at 18 °C.
    2. Calculate the equilibrium concentration of Mg2+ in 1.000 L of saturated MgF2 solution at 18 °C to which 0.100 mol of solid KF has been added. The KF dissolves completely. Assume the volume change is negligible.
    3. Predict whether a precipitate of MgF2 will form when 100.0 mL of a 3.00 × 10–3-M solution of Mg(NO3)2 is mixed with 200.0 mL of a 2.00 × 10–3-M solution of NaF at 18 °C. Show the calculations to support your prediction.
    4. At 27 °C the concentration of Mg2+ in a saturated solution of MgF2 is 1.17 × 10–3 M. Is the dissolving of MgF2 in water an endothermic or an exothermic process? Give an explanation to support your conclusion.

    S15.3.2

    Ksp = [Mg2+][F–]2 = (1.21 × 10–3)(2 × 1.21 × 10–3)2 = 7.09 × 10–9; 7.09 × 10–7 M

    Determine the concentration of Mg2+ and F– that will be present in the final volume. Compare the value of the ion product [Mg2+][F–]2 with Ksp. If this value is larger than Ksp, precipitation will occur. 0.1000 L × 3.00 × 10–3 M Mg(NO3)2 = 0.3000 L × M Mg(NO3)2 M Mg(NO3)2 = 1.00 × 10–3 M 0.2000 L × 2.00 × 10–3 M NaF = 0.3000 L × M NaF M NaF = 1.33 × 10–3 M ion product = (1.00 × 10–3)(1.33 × 10–3)2 = 1.77 × 10–9 This value is smaller than Ksp, so no precipitation will occur. MgF2 is less soluble at 27 °C than at 18 °C. Because added heat acts like an added reagent, when it appears on the product side, the Le Chatelier’s principle states that the equilibrium will shift to the reactants’ side to counter the stress. Consequently, less reagent will dissolve. This situation is found in our case. Therefore, the reaction is exothermic.

    Q15.3.X

    Which of the following compounds, when dissolved in a 0.01-M solution of HClO4, has a solubility greater than in pure water: CuCl, CaCO3, MnS, PbBr2, CaF2? Explain your answer.

    Q15.3.X

    Which of the following compounds, when dissolved in a 0.01-M solution of HClO4, has a solubility greater than in pure water: AgBr, BaF2, Ca3(PO4)3, ZnS, PbI2? Explain your answer.

    BaF2, Ca3(PO4)2, ZnS; each is a salt of a weak acid, and the \(\ce{[H3O+]}\) from perchloric acid reduces the equilibrium concentration of the anion, thereby increasing the concentration of the cations

    Q15.3.X

    What is the effect on the amount of solid Mg(OH)2 that dissolves and the concentrations of Mg2+ and OH– when each of the following are added to a mixture of solid Mg(OH)2 and water at equilibrium?

    1. MgCl2
    2. KOH
    3. HClO4
    4. NaNO3
    5. Mg(OH)2

    Q15.3.X

    What is the effect on the amount of CaHPO4 that dissolves and the concentrations of Ca2+ and \(\ce{HPO4-}\) when each of the following are added to a mixture of solid CaHPO4 and water at equilibrium?

    1. CaCl2
    2. HCl
    3. KClO4
    4. NaOH
    5. CaHPO4

    S15.3.X

    Effect on amount of solid CaHPO4, [Ca2+], [OH–]: increase, increase, decrease; decrease, increase, decrease; no effect, no effect, no effect; decrease, increase, decrease; increase, no effect, no effect

    Q15.3.X

    Identify all chemical species present in an aqueous solution of Ca3(PO4)2 and list these species in decreasing order of their concentrations. (Hint: Remember that the \(\ce{PO4^3-}\) ion is a weak base.)

    Q15.3.X

    A volume of 50 mL of 1.8 M NH3 is mixed with an equal volume of a solution containing 0.95 g of MgCl2. What mass of NH4Cl must be added to the resulting solution to prevent the precipitation of Mg(OH)2?

    S15.3.X

    7.1 g