Acid-Base Equilibria and Solubility Equilibria (산염기 평형 및 용해도 평형) Chapter 16

Acid-Base Equilibria and Solubility Equilibria (산염기 평형 및 용해도 평형) Chapter 16

공통이온효과 (Common ion effect) - The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. ( 공통이온효과란용해된물질과공통인하나의이온을갖는화합물을첨가하여일어나는평형의이동을뜻한다.) - The presence of a common ion suppresses the ionization of a weak acid or a weak base. ( 공통이온의존재는약산또는약염기의이온화를억제한다.) - 예 : Consider mixture of CH 3 COONa (strong electrolyte) and CH 3 COOH (weak acid). ( 강전해질인소듐아세테이트와약산인아세트산의혼합물을고려한다.) CH 3 COONa (s) CH 3 COOH (aq) Na + (aq) + CH 3 COO - (aq) H + (aq) + CH 3 COO - (aq) common ion - 위혼합물의해리반응에서공통이온은아세테이트이온임 - 르샤틀리에원리에의해소듐아세테이트의해리에의해형성된아세테이트이온은아세트산의해리반응의평형을왼쪽으로이동시킴 위혼합액은같은농도의아세트산수용액에비해산의세기가떨어지게됨

- Consider mixture of salt NaA and weak acid HA. ( 어떤염 NaA 와약산 HA 의혼합액을고려한다.) NaA (s) HA (aq) Na + (aq) + A - (aq) H + (aq) + A - (aq) K a = [H+ ][A - ] [HA] [H + ] = K a [HA] [A - ] -log [H + ] = -log K a - log [HA] [A - ] -log [H + ] = -log K a + log [A- ] [HA] ph = pk a + log [A- ] [HA] pk a = -log K a Henderson-Hasselbalch equation ph = pk a + log [conjugate base] [acid] - 헨더슨 - 헨셀발흐식으로부터 K a 와산및염기의농도를알면용액의 ph 를계산가능 (K a 는평형상수이기때문에용액중에산만존재하거나산과그염이공존하는경우에도그값은변하지않음 ) - 헨더슨 - 헨셀발흐식은짝염기의원천과는무관함

교재 677 쪽실전연습 : What is the ph of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? (0.3 M 의아세트산과 0.52 M 의 HCOOK 를포함하는용액의 ph 는얼마인가?) - Mixture of weak acid and conjugate base! ( 약산과짝염기의혼합물에해당함 ) HCOOH (aq) H + (aq) + HCOO - (aq) Initial (M) Change (M) 0.30 0.00 -x +x 0.52 +x Equilibrium (M) 0.30 - x x 0.52 + x Common ion effect 0.30 x 0.30 0.52 + x 0.52 HCOOH pk a = 3.77 ph = pk a + log [HCOO- ] [HCOOH] ph = 3.77 + log [0.52] [0.30] = 4.01

A buffer solution is a solution of: ( 완충용액이란다음에 해당하는용액이다.) 1. A weak acid or a weak base and ( 약산또는약염기 ) 2. The salt of the weak acid or weak base ( 약산또는약염기의염 ) Both must be present! ( 위두성분모두존재해야함 ) - A buffer solution has the ability to resist changes in ph upon the addition of small amounts of either acid or base. ( 완충용액은소량의산이나염기를첨가했을때 ph 의변화에저항하는능력이있음 ) - 완충용액의중요성 : 생물의체액에서 ph 를유지하는데기여함 예 ) 혈액의 ph = 7.4, 위액의 ph = 1.5 - 완충용액의요건 1. 첨가한 OH - 이온과반응할수있도록비교적큰농도의산을포함해야 2. 첨가한 H + 이온과반응하도록비교적큰농도의염기를포함해야 3. 완충제의산과염기들은중화반응에의해서로소비되지않아야함

- Consider an equal molar mixture of CH 3 COOH and CH 3 COONa ( 동일한몰수의아세트산과소듐아세테이트의혼합물을고려한다.) : 소듐아세테이트는강전해질로써, 물에서거의대부분해리될것임 1. Add strong acid ( 강산을첨가하는경우 ) : 소듐아세테이트에서해리된아세테이트이온과반응하여수소이온이소비됨 H + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) 2. Add strong base ( 강염기를첨가하는경우 ) : 아세트산에의해 OH- 이온이중화됨 ) OH - (aq) + CH 3 COOH (aq) CH 3 COO - (aq) + H 2 O (l) 완충용액에산이나염기를첨가할때, 용액의 ph 는거의변하지않고유지될수있음 * 완충용량 (buffer capacity): 완충용액이 ph 를잘유지할수있는능력 ph 가유지되는한계까지첨가된산이나염기의농도로판가름될수있음

* 순수한물과 CH 3 COOH/CH 3 COONa 를포함하는수용액 ( 완충용액 ) 에대해, 동일한양의염산을첨가할때 ph 변화는아래그래프와같다. HCl H + + Cl - HCl + CH 3 COO - CH 3 COOH + Cl - 완충용액에서 ph 는급변하지않고일정하게유지됨

교재 678 쪽실전연습 : Which of the following are buffer systems? ( 다음중완충계는어떤것인가?) (a) KF/HF, (b) KBr/HBr, (c) Na 2 CO 3 /NaHCO 3 (a) HF is a weak acid and F - 짝염기임 ) 완충용액에해당 is its conjugate base (HF 는약산이며 F - 는 (b) HBr is a strong acid (HBr 은강산이기때문에완충용액을형하지않음 ) (c) CO 3 2- is a weak base and HCO 3- is its conjugate acid (CO 3 2- 는약염기이며 HCO 3- 는그짝산임 ) 완충용액에해당 예제 16.3 을통해완충용액의 ph 유지에대해학습해봅시다.

교재 680 쪽실전연습 : Calculate the ph of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the ph after the addition of 20.0 ml of 0.050 M NaOH to 80.0 ml of the buffer solution? (0.3 M 의 NH 3 /0.36 M 의 NH 4 Cl 의 ph 를계산하라. 이완충용액 80 ml 에 0.05 M NaOH 수용액 20 ml 를첨가한다면 ph 는얼마인가?) NH 4 + (aq) H + (aq) + NH 3 (aq) ph = pk a + log [NH 3] [NH 4+ ] pk a = 9.25 ph = 9.25 + log [0.30] [0.36] = 9.17 start (moles) end (moles) 0.029 0.001 0.024 NH + 4 (aq) + OH - (aq) H 2 O (l) + NH 3 (aq) 0.028 0.0 0.025 final volume = 80.0 ml + 20.0 ml = 100 ml [NH 4+ ] = 0.028 0.10 [NH 3 ] = 0.025 0.10 ph = 9.25 + log [0.25] [0.28] = 9.20

Titrations ( 적정 - Review) - In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. ( 적정이란, 정확히농도를알고있는어떤용액에다른미지농도의용액을점진적으로첨가하여두용액사이의화학반응을종료하는과정을지칭한다.) - Equivalence point the point at which the reaction is complete ( 당량점 : 반응이종료되는점을당량점이라부름 ) - Indicator substance that changes color at (or near) the equivalence point ( 지시약 : 당량점이나그부근에서변색된는물질 ) Slowly add base to unknown acid UNTIL The indicator changes color (pink) ( 지시약의색이변할때까지, 미지의산에염기를천천히첨가한다.)

* 복습 : 예제 4.11. What volume of a 1.420 M NaOH solution is required to titrate 25.00 ml of a 4.50 M H 2 SO 4 solution? (4.5 M 의황산수용액 25 ml 에요구되는 1.42 M 의수산화소듐수용액의부피는얼마인가?) 당량점에서황산 1 몰은 NaOH 2 몰과중화반응을완료한다는점을활용할것!!! WRITE THE CHEMICAL EQUATION! H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO 4 M volume acid moles red moles base volume base acid rxn coef. M base ( 중화반응에필요한수산화소듐의몰수 ) 4.50 mol H 2 SO 4 2 mol NaOH 1000 ml soln 25.00 ml x x x = 158 ml 1000 ml soln 1 mol H 2 SO 4 1.420 mol NaOH (25 ml, 4.5 M 에포함된황산의몰수 ) ( 황산 1 몰과수산화소듐 2 몰이반응함 )

당량점의확인방법 * 당량점의요건 : 중화반응을일으키는산과염기의몰수가같을때당량점이됨 * 당량점의확인 : 지시약의변색을통해간접적으로알수있음 * Alternative Method of Equivalence Point Detection ( 당량점을확인하는또다른방법 ): ph 미터를활용하여적정단계에서 ph 변화를직접확인할수있음 ) monitor ph

* Strong Acid-Strong Base Titrations ( 강산과강염기의적정 ) 당량점은 ph = 7 에서발생 NaOH (aq) + HCl (aq) OH - (aq) + H + (aq) H 2 O (l) H 2 O (l) + NaCl (aq)

* Weak Acid-Strong Base Titrations ( 약산과강염기의적정 ) : 중화반응에의해형성된염이가수분해되면서염기로작용 당량점은 ph > 7 에서발생 CH 3 COOH (aq) + NaOH (aq) CH 3 COOH (aq) + OH - (aq) At equivalence point (ph > 7): CH 3 COO - (aq) + H 2 O (l) CH 3 COONa (aq) + H 2 O (l) CH 3 COO - (aq) + H 2 O (l) OH - (aq) + CH 3 COOH (aq)

* Strong Acid-Weak Base Titrations ( 강산과약염기의적정 ) : 중화반응에의해형성된염이가수분해되어산으로작용 당량점에서 ph < 7 이됨 HCl (aq) + NH 3 (aq) NH 4 Cl (aq) H + (aq) + NH 3 (aq) NH 4 Cl (aq) At equivalence point (ph < 7): NH 4 + (aq) + H 2 O (l) NH 3 (aq) + H + (aq)

* 교재 689 쪽실전연습 : Exactly 100 ml of 0.10 M HNO 2 are titrated with a 0.10 M NaOH solution. What is the ph at the equivalence point? ( 정확히 100 ml 의 0.1 M 아질산이 0.1 M 의수산화나트륨에의해적정되었다. (c) 당량점에서 ph 는얼마인가?) 약산 - 강염기적정에해당하며, 635 쪽의표 15.3 의 K b 값참조 Start ( 초기몰수 : moles) End ( 최종몰수 : moles) 0.01 0.01 HNO 2 (aq) + OH - (aq) NO - 2 (aq) + H2 O (l) 0.0 0.0 0.01 Final volume = 200 ml Initial (M) Change (M) Equilibrium (M) K b = [OH- ][HNO 2 ] [NO 2- ] NO 2 - (aq) + H2 O (l) 0.05 0.00 -x +x 0.05 - x = x 2 0.05-x = 2.2 x 10-11 0.05 x 0.05 x 1.05 x 10-6 = [OH - ] [NO 2 - ] = 0.01 0.200 = 0.05 M OH - (aq) + HNO 2 (aq) x poh = 5.98 0.00 +x x ph = 14 poh = 8.02

Acid-Base Indicators ( 산염기지시약 ) - 지시약 (indicator): 이온화되기전후의색이다른약한유기산또는유기염기로써, 산염기적정반응의당량점을확인하는데활용 - 종말점 (termination point): 지시약의색이변하는지점 - HIn: 일양성자성약한산을나타내는기호 ( 지시약을의미 ). 용액에서아래와같이부분적으로이온화된다고가정. HIn (aq) H + (aq) + In - (aq) [HIn] [In - ] [HIn] [In - ] 10 10 Color of acid (HIn) predominates ( 산의색이지배적 ) Color of conjugate base (In - ) predominates ( 염기의색이지배적 )

Solutions of Red Cabbage Extract (적색 양배추 추출물이 들어 있는 용액) ph 18

* The titration curve of a strong acid with a strong base. ( 강산을강염기로적정한곡선 ) - 지시약메틸레드와페놀프탈레인 : 변색되는부분이곡선의급변범위내에있음 적정의당량점을찾는데사용가능 - 티몰블루 : 색깔의변화가적정곡선의급변부분과불일치 : 당량점의확인에사용불가

* 교재 693 쪽실전연습 (c): Which indicator(s) would you use for a titration of HNO 2 with KOH? (HNO 2 와 KOH 의적정에사용가능한지시약은어느것인가?) Weak acid titrated with strong base. At equivalence point, will have conjugate base of weak acid. At equivalence point, ph > 7 Use cresol red or phenolphthalein ( 위반응은약산과강산의반응에해당 당량점은 ph > 7 이됨 지시약의변색범위가 ph > 7 인경우는아래표에서크레졸레드와페놀프탈레인에해당 )

Solubility Equilibria ( 용해도평형 ) * 용해도평형 : 물에서고체침전물이부분적으로이온화되는반응의평형을 용해도평형 이라부름 * 침전반응의예 - 치아에나멜의침식 ( 산성매질에서 ): 하이드록시인회석의용해반응 - 황산바륨의용해도평형 : X 선불투과성화합물 소화기조영제 - 종유석과석순 : 탄산칼슘의침전반응 - 식품의침전반응 : 연한캔디등 Review: 섭씨 25 도의물에대한일반적인이온성화합물의용해규칙

Solubility Equilibria ( 용해도평형 ) * 고체염화은의예 : 물에불용성이지만, 소량의 AgCl은 Ag + 와 Cl - 로이온으로완전해리됨 AgCl (s) K sp = [Ag + ][Cl - ] Ag + (aq) + Cl - (aq) - 위의불균일반응에대한평형상수는다음과같이나타낼수있음 - K sp is the solubility product constant (K sp 를용해도곱상수또는용해도곱이라부름 ) MgF 2 (s) Mg 2+ (aq) + 2F - (aq) K sp = [Mg 2+ ][F - ] 2 Ag 2 CO 3 (s) 2Ag + (aq) + CO 3 2- (aq) Ksp = [Ag + ] 2 [CO 3 2- ] Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq) + 2PO 4 3- (aq) Ksp = [Ca 2+ ] 3 [PO 4 3- ] 2 용해도곱 = 구성이온들의몰농도를곱한것 ( 화학양론계수의멱급수로표시 )

Solubility Equilibria ( 용해도평형 ) * Dissolution of an ionic solid in aqueous solution: ( 수용액에서이온성고체의용해 ) - 용액의불포화, 포화, 과포화의 3 가지상황이존재 - 고체침전물의용해반응이평형에도달하지않았을경우반응지수 Q는이온곱 (ionic product) 라고부름 이온성분의불포화, 과포화를가늠할수있는기준이됨 No precipitate Q < K sp Unsaturated solution ( 불포화용액 ) ( 침전물이존재하지않음 ) Q = K sp Saturated solution ( 포화용액 ) Q > K sp Supersaturated solution ( 과포화용액 ) Precipitate will form ( 침전물이형성됨 ) AgCl (s) Ag + (aq) + Cl - (aq) - Q = [Ag + ] 0 [Cl - ] 0 - K sp = 1.6 X 10-10

섭씨 25 도에서몇몇난용성이온화합물들의용해도곱

몰용해도및용해도 - Molar solubility (mol/l) is the number of moles of solute dissolved in 1 L of a saturated solution. ( 몰용해도는 1 L 의포화용액에녹은용질의몰수로정의됨 ) - Solubility (g/l) is the number of grams of solute dissolved in 1 L of a saturated solution. ( 용해도는 1 L 의포화용액에녹은용질의 g 수로정의됨 ) 용해도자료로부터 K sp 를계산하는과정및 K sp 자료로부터용해도를계산하는과정

* 교재 699 쪽실전연습 : What is the solubility of silver chloride in g/l? (AgCl 의용해도를 g/l 단위로계산하시오 ) AgCl (s) Initial (M) Change (M) -s Equilibrium (M) [Ag + ] = 1.3 x 10-5 M Ag + (aq) + Cl - (aq) 0.00 0.00 +s +s s s [Cl - ] = 1.3 x 10-5 M K sp = 1.6 x 10-10 K sp = [Ag + ][Cl - ] K sp = s 2 s = K sp s = 1.3 x 10-5 Solubility of AgCl = 1.3 x 10-5 mol AgCl 1 L soln x 143.35 g AgCl 1 mol AgCl = 1.9 x 10-3 g/l

K sp 와몰용해도 (s) 사이의관계 27

침전반응의예측 * 두용액을섞을때, 침전형성여부의판별 : 용해도규칙과용해도곱에대한데이터를활용할수있음 * 침전반응의의학적사례 : 신장결석 - 결석은옥살산칼슘으로구성됨 - 혈장중칼슘이온의정상농도 = 5 mm - 시금치등에함유된옥살산이온 : 신체의칼슘과반응하여옥살산칼슘 ( 결석 ) 을형성함 환자의식단조절을통해침전형성을감소시킴

* 교재 700 쪽실전연습 : If 2.00 ml of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl 2, will a precipitate form? ( 1 L 의 0.1 M CaCl 2 용액에 2 ml 의 0.2 M NaOH 를첨가할때침전이형성되겠는가?) - The ions present in solution ( 용액에존재하는이온들 ): Na +, OH -, Ca 2+, Cl -. - Only possible precipitate ( 가능한유일한침전물 ): Ca(OH) 2 (solubility rules). Is Q > K sp for Ca(OH) 2? [Ca 2+ ] 0 = 0.100 M [OH - ] 0 = 4.0 x 10-4 M Q = [Ca 2+ ] 0 [OH - ] 0 2 = 0.10 x (4.0 x 10-4 ) 2 = 1.6 x 10-8 K sp = [Ca 2+ ][OH - ] 2 = 8.0 x 10-6 Q < K sp No precipitate will form ( 불포화상태로, 침전이형성되지않음 )

* 분별침전에의한이온의분리 - 분별침전 (fractional precipitation): 용액안에다른이온을남겨둔채한종류의이온을침전으로제거하는과정을칭함 - 분별침전의예 : K + 와 Ba 2+ 를포함하는용액으로부터황산이온을첨가하여 BaSO 4 를침전시키는과정 (K 2 SO 4 는가용성이므로용액에남아있을것임 ) - Cl-, Br-, I- 를포함하는용액을고려 : 질산은과같은가용성화합물을첨가하면 AgI, AgBr, AgCl 순서로침전됨 ( 교재 700 쪽좌측표참조 )

* 예제 16.11: What concentration of Ag is required to precipitate ONLY AgBr in a solution that contains both Br - and Cl - at a concentration of 0.02 M? (0.02 M 의 Br- 와 Cl- 를모두포함하는용액에서 AgBr 만을침전시키기위하여필요한 Ag 의농도는얼마이가?) AgBr (s) Ag + (aq) + Br - (aq) K sp = 7.7 x 10-13 K sp = [Ag + ][Br - ] [Ag + ] = K sp [Br - ] 7.7 x 10 = -13 = 3.9 x 10 0.020-11 M AgCl (s) [Ag + ] = Ag + (aq) + Cl - (aq) K sp [Cl - ] K sp = 1.6 x 10-10 K sp = [Ag + ][Cl - ] 1.6 x 10 = -10 = 8.0 x 10 0.020-9 M AgCl AgBr 이농도는 AgCl 까지침전되기시작하는 Ag 의농도임 따라서 AgBr 만침전되기위한 Ag 의농도는다음범위에해당함 3.9 x 10-11 M < [Ag + ] < 8.0 x 10-9 M

The Common Ion Effect and Solubility ( 공통이온효과와용해도 ) - The presence of a common ion decreases the solubility of the salt. ( 공통이온의존재는염의용해도를감소시킴 ) 그이유를설명하기위하여다음예를생각해보자. 예 ) AgNO 3 와 AgCl 이함께용해된용액을고려함 AgNO 3 (s) Ag + + NO 3- ( 비가역반응 ) AgCl (s) Ag + + Cl - ( 평형반응 ) (Ag+ 는공통이온으로작용함 ) - 용액중 Ag + 이온의농도증가 르샤틀리에원리에의해 AgCl 의용해반응은평형이왼쪽으로이동 일부의이온이 AgCl 로침전됨 염의용해도는감소하는효과

The Common Ion Effect and Solubility ( 공통이온효과와용해도 ) * 교재 703 쪽실전연습 : What is the molar solubility of AgBr in (a) pure water and (b) 0.0010 M NaBr? (a. 순수한물과 b. 0.001 M NaBr 수용액에서 AgBr 의몰용해도는얼마인가?) AgBr (s) K sp = 7.7 x 10-13 s 2 = K sp s = 8.8 x 10-7 Ag + (aq) + Br - (aq) NaBr (s) Na + (aq) + Br - (aq) [Br - ] = 0.0010 M AgBr (s) Ag + (aq) + Br - (aq) [Ag + ] = s [Br - ] = 0.0010 + s 0.0010 K sp = 0.0010 x s s = 7.7 x 10-10

ph and Solubility (ph 및용해도 ) The presence of a common ion decreases the solubility. ( 공통이온의존재는용해도를감소시킴 ) 르샤틀리에의원리에의해화학평형이왼쪽으로이동하며염의농도증가 / 용해도감소 Insoluble bases dissolve in acidic solutions ( 불용성염기는산용액에녹음 ) Insoluble acids dissolve in basic solutions ( 불용성산은염기용액에녹음 ) Mg(OH) 2 (s) remove add Mg 2+ (aq) + 2OH - (aq) K sp = [Mg 2+ ][OH - ] 2 = 1.2 x 10-11 K sp = (s)(2s) 2 = 4s 3 4s 3 = 1.2 x 10-11 s = 1.4 x 10-4 M [OH - ] = 2s = 2.8 x 10-4 M poh = 3.55 ph = 10.45 At ph less than 10.45 Lower [OH - ] OH - (aq) + H + (aq) H 2 O (l) Increase solubility of Mg(OH) 2 At ph greater than 10.45 Raise [OH - ] Decrease solubility of Mg(OH) 2

Complex Ion Equilibria and Solubility ( 착이온평형및용해도 ) - A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. ( 착이온이란중심금속양이온이하나이상의분자나 이온과결합한형태로정의됨 ) Co 2+ (aq) + 4Cl - (aq) K f = 2- [CoCl 4 ] [Co 2+ ][Cl - ] 4 2- CoCl 4 (aq) - The formation constant or stability constant (K f ) is the equilibrium constant for the complex ion formation. ( 착이온형성반응에대한평형상수를형성상수또는안정도상수 K f 라정의함 ) - 위의착이온형성반응에대한형성상수는다음과같이정의할수있음 Co(H 2 O) 6 2+ CoCl 4 2- HCl K f stability of complex - 안정도상수값이커지면착이온이더잘형성되며착이온의안정도는증가함

Effect of Complexation on Solubility ( 용해도에대한착이온형성의효과 ) AgNO Add 3 + NH NaCl 3 * 착이온이형성되면염의용해도는증가되는효과가얻어짐 예제 16.16. 1M 의 NH 3 용액에대한 AgCl 의용해도를계산하는문제 Ag(NH AgCl 3 ) + 2

몇가지착이온의형성상수 37

정성분석에서용해도곱원리의응용 - 정성분석 (qualitative analysis): 어떤성분의존재여부를확인하는분석에해당 - 정량분석 (quantative analysis): 어떤성분이얼마나많이존재하는지판단하는분석에해당 - 수용액중에존재를확인할수있는양이온은약 20 가지정도 ( 정성분석이가능 ) 몇가지침전재를첨가하여차례로확인가능 - 침전되는순서에따라양이온들은 1 족부터 5 족까지로분류함 ( 주기율표의원소들을족으로분류하는체계와다름, 표 16.5 참조 )

여러가지시약과의침전반응에따라양이온을족으로분류 39

Qualitative Analysis of Cations ( 정성분석에서양이온을분리하는계통도 ) 나머지이온들이존재하는용액 + NaOH 여과 3 족은 CoS, FeS, MnS, NiS, ZnS, Al(OH) 3, Cr(OH) 3 의침전이발생 5 족은불꽃테스트등으로확인함

Flame Test for Cations ( 양이온의불꽃테스트 ) - 5 족양이온 : 정성분석에서마지막에남는물질은 5 족양이온에해당 ( 나트륨양이온, 칼륨양이온, 암모늄양이온 ) - 암모늄양이온 : 냄새로확인가능하나붉은리트머스시험지를푸르게변화시킴 - 나트륨양이온 : 불꽃테스트로존재를확인가능 ( 백금선조각에용액을적시고분젠버너불꽃으로가열하면노란색을띔 ) - 칼륨양이온 : 불꽃테스트에서보라색을띔 Lithium ( 붉은색 ) Sodium ( 노랑 ) Potassium ( 보라 ) Copper ( 초록 )

Chemistry In Action: How an Eggshell is Formed ( 생활속의화학 : 달걀껍질은어떻게형성되는가 ) - 달걀껍질은칼슘양이온이탄산음이온과다음과같은침전반응을일으켜형성됨 Ca 2+ (aq) + CO 3 2- (aq) CO 2 (g) + H 2 O (l) H 2 CO 3 (aq) HCO 3- (aq) CaCO 3 (s) - 닭의체내에서이산화탄소는탄산탈수효소 (carbonic anhydrase) 의작용에의해탄산으로변함 carbonic H 2 CO 3 (aq) anhydrase - 탄산은단계적인이온화를거쳐탄산음이온을형성함 H + (aq) + HCO 3- (aq) H + (aq) + CO 3 2- (aq) electron micrograph

Chemistry In Action: How an Eggshell is Formed ( 생활속의화학 : 달걀껍질은어떻게형성되는가 ) - 계란껍질의칼슘은닭의뼈에서유래됨 - 닭은한개의계란을낳기위해뼈의칼슘을 10 % 소모함 닭의모이에포함된칼슘의양이매우적을경우계란을낳을수없음 - 더운날닭이헐떡거림 닭의체내에존재하는이산화탄소가배출됨 르샤틀리에의원리에의해다음반응의평형이왼쪽으로이동 달걀껍질의원료인탄산의농도가감소 껍질이얇은계란을낳게됨 CO 2 (g) + H 2 O (l) carbonic H 2 CO 3 (aq) anhydrase - 해결방법 : 더운날닭에게탄산수를먹인다.

Chemistry In Action: Maintaining the ph of Blood Red blood cells in a capillary 44