Baixe Resolução de exercicios do livro Skoog 8ª edição e outras Resumos em PDF para Química Analítica, somente na Docsity!
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8 thth ed.ed. ChapterChapter 44 Chapter 4Chapter 4 4-14-1 (a)(a) TheThe millimolemillimole is an amount of a chemical species, such as an atom, an ion, a moleculeis an amount of a chemical species, such as an atom, an ion, a molecule oror anan electron.electron. AA millimolmillimolee containscontains millimolemillimole particlesparticles 66 .. 0202 1010 millimolemillimole molemole 1010 molemole particlesparticles 66 .. 0202 1010 2323 33 2020 ×× ∗∗ == ×× −− (b)(b) TheThe molar massmolar mass is the mass in grams of one mole of a chemical species.is the mass in grams of one mole of a chemical species. (c)(c) TheThe millimolar massmillimolar mass is the mass in grams of one millimole of a chemical species.is the mass in grams of one millimole of a chemical species. (d)(d) Parts per million, cParts per million, cppmppm, is a term expressing the concentration of very dilute solutions., is a term expressing the concentration of very dilute solutions. Thus,Thus, ccppmppm 1010 ppmppm massmassofof solutionsolution massmass ofof solutesolute 66 == ×× The units of mass in the numerator and the denominator must be the same.The units of mass in the numerator and the denominator must be the same. 4-24-2 The species molarity of a solution expresses the equilibrium concentration of a chemicalThe species molarity of a solution expresses the equilibrium concentration of a chemical species in terms ofspecies in terms of moles per liter.moles per liter. The analytical molarity of aThe analytical molarity of a solution gives the totalsolution gives the total number of moles ofnumber of moles of a solute in onea solute in one liter.liter. The species molarity takes into accountThe species molarity takes into account chemical reactions that occurchemical reactions that occur in solution.in solution. The analytical molarity specifies howThe analytical molarity specifies how thethe solution was prepared, but does not account for any subsequent reactions.solution was prepared, but does not account for any subsequent reactions. 4-34- 33 33 33 33 1010 mm 100100 cmcm mm mLmL 11 cmcm 11 LL 10001000 mLmL 1L1L −− ==
== ×× ×× 33 33 33 33 1010 mm 11 molemole 1010 mm
LL
LL
11 molemole 11 MM == ×× −− == −−
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8 thth ed.ed. ChapterChapter 44 4-84-8 (a)(a) (^2255) 22 55 22 55 22 55 00 ..^4040 mmolmmolPPOO molmol 10001000 mmolmmol 141141 .. 9494 ggPP OO molmolPP OO 10001000 mgmg gg 5757 mgmgPP OO ×× ×× ×× == (b)(b) (^22) 22 22 22 293293 ..^66 mmolmmolCOCO molmol 10001000 mmol mmol 4444 .. 0101 gg COCO molmol COCO 1212 .. 9292 gg COCO ×× ×× == (c)(c) (^33) 33 33 33 476476 mmolmmolNaHCONaHCO molmol 10001000 mmolmmol 8484 .. 0101 gg NaHCONaHCO molmol NaHCONaHCO 4040 .. 00 gg NaHCONaHCO ×× ×× == (d)(d) 44 44 44 44 44 44 44 44 66 .. 22 mmolmmolMgNHMgNH POPO molmol 10001000 mmolmmol 137.32137.32ggMgNHMgNH POPO molmolMgNHMgNH POPO 10001000 mgmg gg 850850 mgmgMgNHMgNH POPO == ×× ×× ×× 4-94-9 (a)(a) 44 44 33 44 33 66 .. 5050 mmolmmolKMnO KMnO
22 .. 0000 LL
molmol 10001000 mmolmmol LL 33 .. 2525 1010 molmol KMnOKMnO 33 .. 2525 1010 M KMnOM KMnO == ×× ×× ×× ×× ≡≡ −− −− (b)(b) 4141 .. 66 mmolmmol KSCNKSCN 750750 mLmL 10001000 mLmL
LL
molmol 10001000 mmolmmol LL 00 .. 05550555 molmol KSCNKSCN 00 .. 05550555 M KSCNM KSCN == ≡≡ ×× ×× ×× (c)(c) 44 33 44 44 44 44 250250 mLmL 88 .. 4747 1010 mmolmmol CuSOCuSO 10001000 mLmL
LL
molmol 10001000 mmolmmol 159159 .. 6161 gg CuSOCuSO molmol CuSOCuSO 10001000 mgmg gg LL 55 .. 4141 mgmg CuSOCuSO 55 .. 4141 ppmppm CuSOCuSO −− ×× ×× == ×× ≡≡ ×× ×× ×× (d)(d) 33 .. 5050 LL 11651165 .. 66 mmolmmolKClKCl molmol 10001000 mmolmmol LL 00 .. 333333 molmol KClKCl 00 .. 333333 M KClM KCl≡≡ ×× ×× ==
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8 thth ed.ed. ChapterChapter 44 4-104-10 (a)(a) 44 44 44 5656 .. 00 mmolmmolHClO HClO 175175 mLmL 10001000 mLmL
LL
molmol 10001000 mmolmmol LL 00 .. 320320 molmol HClOHClO 00 .. 320320 MM HClOHClO == ≡≡ ×× ×× ×× (b)(b) 22 44 22 44 33 22 44 33 121121 mmolmmolKK CrOCrO
1515 .. 00 LL
molmol 10001000 mmolmmol LL 88 .. 0505 1010 molmolKK CrOCrO 88 .. 0505 1010 MMKK CrOCrO == ×× ×× ×× ×× ≡≡ −− −− (c)(c) 33 33 33 33 33 55 .. 0000 LL 00 .. 199199 mmolmmol AgNOAgNO molmol 10001000 mmolmmol 169169 .. 8787 gg AgNOAgNO molmol AgNOAgNO 10001000 mgmg gg LL 66 .. 7575 mgmg AgNOAgNO 66 .. 7575 ppmppm AgNOAgNO == ≡≡ ×× ×× ×× ×× (d)(d) 1717 .. 00 mmolmmol KOHKOH 851851 mLmL 10001000 mLmL
LL
molmol 10001000 mmolmmol LL 00 .. 02000200 molmol KOHKOH 00 .. 02000200 MM KOHKOH == ≡≡ ×× ×× ×× 4-114-11 (a)(a) (^33) 44 33 33 33 44 ..^90901010 mgmgHNOHNO gg 10001000 mgmg molmol HNOHNO 6363 .. 0101 gg HNOHNO 00 .. 777777 molmolHNO HNO ×× ×× == ×× (b)(b) 22 .. 015015 1010 mgmgMgOMgO gg 10001000 mgmg molmol MgOMgO 40.3040.30gg MgOMgO 10001000 mmolmmol molmol 500500 mmolmmol MgOMgO 44 ×× ×× ×× == ×× (c)(c) (^4433) 66 44 33 44 33 44 33 11 ..^80801010 mgmgNHNH NONO gg 10001000 mgmg molmolNHNH NONO 8080 .. 0404 ggNHNH NONO 2222 .. 55 molmolNHNH NONO ×× ×× == ×× (d)(d) 44 22 33 66 66 44 22 33 66 44 22 33 66 44 22 33 66 22 .. 3737 1010 mgmg((NHNH )) CeCe((NONO )) gg 10001000 mgmg molmol((NHNH )) CeCe((NONO )) 548548 .. 2323 gg((NHNH )) CeCe((NONO )) 44 .. 3232 molmol((NHNH )) CeCe((NONO )) == ×× ×× ××
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8 thth ed.ed. ChapterChapter 44 4-144-14 (a)(a) 22 22 22 22 22 22 22 22 22 22 22 .. 5151 ggHH OO 450450 mLmL molmolHH OO 34.0234.02ggHH OO 10001000 mLmL
LL
LL
0.1640.164molmolHH OO 00 .. 164164 MMHH OO == ≡≡ ×× ×× ×× (b)(b) 2727 .. 00 mLmL 22 .. 8888 1010 ggbenzoicbenzoic acidacid molmolbenzoicbenzoic acidacid 122122 ggbenzoicbenzoic acidacid 10001000 mLmL
LL
LL
88 .. 7575 1010 molmolbenzoicbenzoic acidacid 88 .. 7575 1010 MMbenzoicbenzoic acidacid 33 44 44 −− −− −− == ×× ×× ×× ×× ×× ×× ≡≡ (c)(c) (^22) 22 22 33 ..^5050 LL^00 ..^07600760 ggSnClSnCl 10001000 mgmg gg LL 2121 .. 77 mgmg SnClSnCl 2121 .. 77 ppmppm SnClSnCl ≡≡ ×× ×× == (d)(d) 33 33 33 33 33 00 .. 04530453 gg KBrOKBrO 2121 .. 77 mLmL molmol KBrOKBrO 167167 gg KBrOKBrO 10001000 mLmL
LL
LL
00 .. 01250125 molmol KBrOKBrO 00 .. 01250125 MM KBrOKBrO == ≡≡ ×× ×× ×× 4-154-15 (a)(a) 00 .. 923923 (( 22 )) 11 .. 077077 log(log( 88 .. 3838 )) log(log( 1010 )) pNapNa log(log( 00 .. 03350335 MM 00 .. 05030503 MM)) log(log( 00 .. 08380838 MM)) log(log( 88 .. 3838 1010 MM )) 22 22 ==−− −− −− == ==−− −− ==−− ++ ==−− ==−− × × −− −− 00 .. 525525 (( 22 )) 11 .. 475475 log(log( 33 .. 3535 )) log(log( 1010 )) pClpCl log(log( 00 .. 03350335 MM)) log(log( 33 .. 3535 1010 MM )) 22 22 ==−− −− −− == ==−− −− ==−− ==−− ×× −− −− 00 .. 702702 (( 22 )) 11 .. 298298 log(log( 55 .. 0303 )) log(log( 1010 )) pOHpOH log(log( 00 .. 05030503 MM)) log(log( 55 .. 0303 1010 MM )) 22 22 ==−− −− −− == ==−− −− ==−− ==−− ×× −− − −
Fundamentals of Analytical Chemistry: 8 th ed. Chapter 4 (b)
- 884 ( 3 ) 2. 116 log( 7. 65 ) log( 10 ) pBa log( 7. 65 10 M ) 3 3 =− − − = =− − =− × − − pMn =−log( 1. 54 M)=− 0. 188 pCl log( 7. 65 10 M ( 2 1. 54 M)) log( 3. 08 M) 0. 490 3 = − × + × =− =− − (c)
- 778 ( 1 ) 0. 222 log( 6. 00 ) log( 10 ) pH log( 0. 600 M) log( 6. 00 10 M ) 1 1 =− − − = =− − =− =− × − −
- 904 ( 1 ) 0. 096 log( 8. 02 ) log( 10 ) pCl log( 0. 600 M ( 2 0. 101 M)) log( 0. 802 M) log( 8. 02 10 M ) 1 1 = − − = =− − =− + × =− =− × − −
- 00432 ( 1 ) 0. 996 log( 1. 01 ) log( 10 ) pZn log( 0. 101 M) log( 1. 01 10 M ) 1 1 =− − − = =− − =− =− × − − (d)
- 679 ( 2 ) 1. 320 log( 4. 78 ) log( 10 ) pCu log( 4. 78 10 M ) 2 2 =− − − = =− − =− × − −
- 0170 ( 1 ) 0. 983 log( 1. 04 ) log( 10 ) pZn log( 0. 104 M) log( 1. 04 10 M ) 1 1 =− − − = =− − =− =− × − −
- 483 ( 1 ) 0. 517 log( 3. 04 ) log( 10 ) pNO log(( 2 0. 0478 M) ( 2 0. 104 M)) log( 0. 304 M) log( 3. 04 10 M ) 1 1 3 =− − − = =− − =− × + × =− =− × − −
Fundamentals of Analytical Chemistry: 8 th ed. Chapter 4 (d) [ HO ] 2. 4 10 M 14 3
- − = × (e) [H O ] 4. 8 10 M 8 3
- − = × (f) [H O ] 1. 7 10 M 6 3
- − = × (g) [ H 3 O ]= 2. 04 M
(h) [ H 3 O ]= 3. 3 M
4-17 (a)
- 301 ( 2 ) 1. 699 pNa pBr log( 0. 0200 M) log( 2. 00 ) log( 10 ) 2 =− − − = = =− =− − − − pH = pOH = - log(1.0× 10
- 000 ( 2 ) 2. 000 pBa log( 0. 0100 M) log( 1. 00 ) log( 10 ) 2 = − − = = − =− − −
- 301 ( 2 ) 1. 699 pBr log( 2 0. 0100 M) log( 2. 00 ) log( 10 ) 2 =− − − = = − × =− − − pH = pOH = - log(1.0× 10
M) = 7. (c)
- 54 ( 3 ) 2. 46 pBa log( 3. 5 10 M) log( 3. 5 ) log( 10 ) 3 3 =− − − = = − × =− − − −
- 84 ( 3 ) 2. 15 pOH log( 2 ( 3. 5 10 M) log( 7. 0 10 M) log( 7. 0 ) log( 10 ) 3 3 3 =− − − = = − × × =− × =− − − − − pH = 14.00 – pOH = (14.00 – 2.15) = 11.
Fundamentals of Analytical Chemistry: 8 th ed. Chapter 4 (d)
- 60 ( 2 ) 1. 40 pH log( 0. 040 M) log( 4. 0 10 M) log( 4. 0 ) log( 10 ) 2 2 =− − − = = − =− × =− − − −
- 30 ( 2 ) 1. 70 pNa log( 0. 020 M) log( 2. 0 10 M) log( 2. 0 ) log( 10 ) 2 2 =− − − = = − =− × =− − − −
- 78 ( 2 ) 1. 22 log( 6. 0 ) log( 10 ) pCl log( 0. 040 M 0. 020 M) log( 0. 060 M) log( 6. 0 10 M ) 2 2 =− − − = =− − =− + =− =− × − − pOH = 14.00 – 1.40 = 12. (e)
- 83 ( 3 ) 2. 17 pCa log( 6. 7 10 M) log( 6. 7 ) log( 10 ) 3 3 =− − − = = × =− − − −
- 88 ( 3 ) 2. 12 pBa log( 7. 6 10 M) log( 7. 6 ) log( 10 ) 3 3 =− − − = = − × =− − − −
- 46 ( 2 ) 1. 54 log( 2. 9 ) log( 10 ) pCl log(( 2 ( 6. 7 10 M)) ( 2 ( 7. 6 10 M))) log( 2. 9 10 M ) 2 3 3 2 =− − − = =− − =− × × + × × =− × − − − − pH = pOH = - log(1.0× 10
- 68 ( 8 ) 7. 32 pZn log( 4. 8 10 M) log( 4. 8 ) log( 10 ) 8 8 =− − − = = − × =− − − −
- 75 ( 7 ) 6. 25 pCd log( 5. 6 10 M) log( 5. 6 ) log( 10 ) 7 7 =− − − = = − × =− − − −
Fundamentals of Analytical Chemistry: 8 th ed. Chapter 4
- 458 ( 3 ) 2. 543 pSO log( 2. 87 10 M) log( 2. 87 ) log( 10 ) 3 3 4 =− − − = = − × =− − − − 4-20 (a) − +
× × × = 4. 6 × 10 MK
- 10 g mol K 1000 mg g L 1000 mL 100 mL 18 mg K 3 − − − × × × = 0. 103 MCl
- 45 g mol Cl 1000 mg g L 1000 mL 100 mL 365 mg Cl (b)
- 66 ( 3 ) 2. 34 pK log( 4. 6 10 M) log( 4. 6 ) log( 10 ) 3 3 =− − − = = − × =− − − − pCl = –log(1.03× 10
- M) = –log(1.03)–log( - ) = –0.0133–(–1) = 0. 4-21 (a)
- 01037 MKCl MgCl 6 H O
- 85 g molKCl MgCl 6 H O
- 00 L
- 76 gKCl MgCl 6 H O 2 2 2 2 2 2 = ⋅ ⋅ ⋅ ⋅ × ⋅ ⋅ (b)
= ⋅ ⋅ ⋅ ⋅ × 2 2 2 2 2 2 0.^01037 MMg molKCl MgCl 6 H O mol Mg
- 01037 MKCl MgCl 6 H O (c) − − = ⋅ ⋅ ⋅ ⋅ × 0. 0311 MCl molKCl MgCl 6 H O 3 mol Cl
- 01037 MKCl MgCl 6 H O 2 2 2 2 (d) 100 % 0. 288 %(w/v) 1000 mL
L
2. 00 L
- 76 gKCl MgCl 2 6 H 2 O × × = ⋅ ⋅
Fundamentals of Analytical Chemistry: 8 th ed. Chapter 4 (e) − − − ≡ × × ×25.0mL = 0. 777 mmolCl mol 1000 mmol 1000 mL
L
L
0.0311mol Cl
- 0311 M Cl (f)
= ≡ × × ⋅ ⋅ ⋅ ⋅ × 405 ppm K L 405 mg g 1000 mg mol K
- 10 g K 1 molKCl MgCl 6 H O 1 mol K
- 01037 MKCl MgCl 6 H O 2 2 2 2 (g) pMg log( 1. 04 10 M) log( 1. 04 ) log( 10 ) 0. 0170 ( 2 ) 1. 984 2 2 = − × =− − =− − − = − − (h) pCl^ log(^3.^1210 M) log(^3.^12 ) log(^10 )^0.^494 (^2 )^1.^507 2 2 = − × =− − =− − − = − − 4-22 (a) 3 6 3 6 3 6 3 6 3
- 74 10 MK Fe(CN)
- 2 g molK Fe(CN ) 775 L 1210 gK Fe(CN ) 775 mL 1210 mgK Fe(CN ) − ≡ × = × (b)
− × × = 0. 0142 MK molK Fe(CN ) 3 mol K
- 74 10 MK Fe(CN ) 3 6 3 6 3 (c) − − − − × × = × 3 6 3 3 6 3 6 3 6 3
- 74 10 MFe(CN) molK Fe(CN ) molFe(CN )
- 74 10 MK Fe(CN ) (d) 100 % 0. 156 %(w/v) 1000 mg g 775 mL 1210 mgK 3 Fe(CN ) 6 × × = (e)
− + + − + = × × × × × ≡
- 710 mmol K 50.0 mL mol K 1000 mmolK 1000 mL
L
L
1.42 10 mol K
- 42 10 M K 2 2
Fundamentals of Analytical Chemistry: 8 th ed. Chapter 4 4-25 (a) 500 mLsoln 23. 8 gC H OH 100
mL soln
- 75 gC H OH
- 75 %(w/v)C 2 H 5 OH≡^25 × × = 2 5 Weigh 23.8 g ethanol and add enough water to give a final volume of 500 mL. (b) gwater 500 gsoln 23. 8 gC HOH 476. 2 g water 500 gsoln 23.8gC HOH g water 500 gsoln 23. 8 gC H OH 100
g soln
- 75 gC H OH
- 75 %(w/w)C H OH 2 5 2 5 2 5 2 5 2 5 = − = = + ≡ × × = x x Mix 23.8 g ethanol with 476.2 g water. (c) 500 mLsoln 23. 8 mLC H OH 100
mL soln
- 75 mLC H OH
- 75 %(v/v)C H OH 2 5 2 5 2 5 ≡ × × = Dilute 23.8 mL ethanol with enough water to give a final volume of 500 mL. 4-26 (a) 3 8 3 3 8 3 3 8 3 2.^50 Lsoln^525 gC HO L 1000 mL 100
mL soln
- 0 gC H O
- 0 %(w/v)C H O ≡ × × × = Weigh 525 g glycerol and add enough water to give a final volume of 2.50 L. (b) kgwater 2. 50 kgsoln 0. 525 kgCHO 1. 98 kg water
- 50 kgsoln 0.525kgCHO kg water
- 50 kgsoln 525 gC H O kg 1000 g 100
g soln
- 0 gC H O
- 0 %(w/w)C H O 3 8 3 3 8 3 3 8 3 3 8 3 3 8 3 = − = = + ≡ × × × = x x Mix 525 g glycerol with 1.98 kg water.
Fundamentals of Analytical Chemistry: 8 th ed. Chapter 4 (c) 3 8 3 3 8 3 3 8 3 2.^50 Lsoln^525 mLC HO L 1000 mL 100
mL soln
- 0 mLCH O
- 0 %(v/v)CH O ≡ × × × = Dilute 525 mL glycerol with enough water to give a final volume of 2.50 L. 4-
- 300 L
- 0 molH PO
L
volume 86 %(w/w)H PO required 4. 50 molH PO
L
- 0 molH PO
- 0 g molH PO L 1000 mL mL g water g water
- 71 g reagent 100
g reagent 86 gH PO 86 %(w/w)H PO 750 mL 4. 50 molH PO 1000 mL
L
L
- 00 molH PO
- 00 MH PO 3 4 3 4 3 4 3 4 3 4 3 4 3 4 3 4 3 4 3 4 = × = = ≡ × × × × × ≡ × × = Dilute 300 mL of the concentrated reagent to 750 mL using water. 4-
- 170 L
- 9 mol HNO
L
volume 70. 5 %HNO required 2. 70 mol HNO
L
- 9 mol HNO
- 0 g mol HNO L 1000 mL mL g water g water
- 42 g reagent 100
g reagent
- 5 g HNO
- 5 %(w/w) HNO 900 mL 2. 70 mol HNO 1000 mL
L
L
- 00 mol HNO
- 00 M HNO 3 3 3 3 3 3 3 3 3 3 = × = = ≡ × × × × × ≡ × × = Dilute 170 mL of the concentrated reagent to 900 mL using water. 4-29 (a) 3 3 3 3
- 37 g AgNO 500 mL 1000 mL
L
mol
- 87 g AgNO L
- 0750 mol AgNO
- 0750 M AgNO = ≡ × × × Dissolve 6.37 g AgNO 3 in enough water to give a final volume of 500 mL.
Fundamentals of Analytical Chemistry: 8 th ed. Chapter 4 (f) 2 4 2 4 2 4
- 67 gNa SO mol
- 04 gNa SO 2 mol Na molNa SO
- 02348 mol Na
- 02348 mol Na
- 99 g mol Na 1000 mg g 540 mg Na
- 00 L 540 mg Na L soln 60 mg Na 60 ppm Na × × = × × = ≡ × =
Dissolve 1.67 g Na 2 SO 4 in enough water to give a final volume of 9.00 L. 4-30 (a) 4 4 4 4 39.^5 gKMnO mol
- 03 g KMnO
- 00 L L
- 0500 mol KMnO
- 0500 M KMnO ≡ × × = Dissolve 39.5 g KMnO 4 in enough water to give a final volume of 5.00 L. (b)
- 125 L reagent
- 00 mol reagent
L
- 00 mol HClO
- 00 L 1. 00 mol HClO L
- 250 mol HClO
- 250 M HClO 4 4 4 4 × = ≡ × = Take 125 mL of the 8.00 M reagent and dilute a final of volume of 4.00 L using water. (c) 2 (^2 221). 39 gMgI mol
- 11 g MgI 2 mol I mol MgI
- 00 10 mol I 400 mL 0. 0100 mol I 1000 mL
L
L
- 0250 mol I
- 0250 M I × × × = ≡ × × = − − − − − − Dissolve 1.39 g MgI 2 in enough water to give a final volume of 400 mL. (d)
Fundamentals of Analytical Chemistry: 8 th ed. Chapter 4
- 0343 L
- 365 mol CuSO
L
- 60 g mol CuSO
- 00 g CuSO 200 mL 2. 00 g CuSO 100
mL soln
- 00 g CuSO
- 00 %(w/v) CuSO 4 4 4 4 4 4 × × = ≡ × × = Take 34.3 mL of the 0.365 M CuSO 4 and dilute to a final volume of 200 mL using water. (e)
- 0169 L
- 062 mol NaOH
L
volume 50 %(w/w)NaOHrequired 0. 3225 mol NaOH
L
- 062 mol NaOH
- 00 g mol NaOH L 1000 mL mL g water g water
- 525 g reagent 100
g reagent 50 g NaOH 50 %(w/w) NaOH
- 50 L 0. 3225 mol NaOH L
- 215 mol NaOH
- 215 M NaOH = × = = ≡ × × × × × ≡ × = Take 16.9 mL of the concentrated reagent and dilute to a final volume of 1.50 L using water. (f) 4 6 (^44646) 1 4 1
- 0424 gK Fe(CN ) mol
- 35 gK Fe(CN ) 4 mol K molK Fe(CN )
- 60 10 mol K
- 60 10 mol K
- 10 g mol K 1000 mg g
- 8 10 mg K
- 50 L 1. 8 10 mg K L soln 12 mg K
- 0 ppm K × × × = × × × = × ≡ × = ×
− + − +
Dissolve 42.4 mg K 4 Fe(CN) 6 in enough water to give a final volume of 1.50 L. 4- −^ − − − − +
≡ × × = × ≡ × × = × 3 3 2 3 2 3 3 3
- 0 mL 2. 27 10 mol IO 1000 mL
L
L
- 302 mol IO
- 302 M IO
- 0 mL 1. 25 10 mol La 1000 mL
L
L
- 250 mol La
- 250 M La Because each mole of La(IO 3 ) 3 requires three moles IO (^3)
- , IO 3 - becomes the limiting
