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Chapter 17: Aqueous Ionic Equilibrium

Alexandra Bozan General Chemistry Inglês 106 flashcards

calculate the pH of a buffer that is 0.200 M in HC₂H₃O₂ and 0.100 M in NaC₂H₃O₂. HC₂H₃O₂ Kₐ = 1.8 × 10⁻⁵ step 2: Use Ice…

Flashcards do conjunto (106)

  1. 1buffers- solutions that resist pH change
    - contains weak acid and conjugate base or weak base and conjugate acid
    weak acid = HNO₂ weak base = NaNO₂
  2. 2solubility equilibriaextent to which slightly soluble ionic compounds dissolve in water
  3. 3weak acid or base by itself doesn't contain sufficient _____ or _____ to be a bufferbase
    acid
  4. 4to determine the pH of a buffer solution containing common ions, the initial [] should include the _____ and its _____ _____acid
    conjugate base
  5. 5calculate the pH of a buffer that is 0.200 M in HC₂H₃O₂ and 0.100 M in NaC₂H₃O₂. HC₂H₃O₂ Kₐ = 1.8 × 10⁻⁵ step 1: write the balanced equation for ionization of the acidstep 1: write the balanced equation for ionization of the acid
    HC₂H₃O₂ + H2O ⇆ H₃O⁺ + C₂H₃O₂⁻
  6. 6calculate the pH of a buffer that is 0.200 M in HC₂H₃O₂ and 0.100 M in NaC₂H₃O₂. HC₂H₃O₂ Kₐ = 1.8 × 10⁻⁵ step 2: Use Ice for weak acid solutionscalculate the pH of a buffer that is 0.200 M in HC₂H₃O₂ and 0.100 M in NaC₂H₃O₂. HC₂H₃O₂ Kₐ = 1.8 × 10⁻⁵ step 2: Use Ice for weak acid so...
  7. 7to determine [H₃O⁺] for any buffer solution, multiply Kₐ by the ratio of the [] of the _____ and _____ _____acid
    conjugate base
    [H₃O⁺] = Kₐ × ([HA]/[A⁻])
  8. 8log AB = log A + log B then -log AB =-log A - log B
  9. 9in any solution where acid/conjugate base [] are = then [H₃O⁺] = _____ and the pH of the buffer is = _____Kₐ
    pKₐ
  10. 10Henderson-Hasselbalch equationallows us to quickly calculate the pH of a buffer solution from the initial [] of the buffer components (as long as x is small approximation)
  11. 11Henderson-Hasselbalch equation formulaEnter text here...Henderson-Hasselbalch equation formula
  12. 12If the pH of the buffer is < than pKₐ of the acid, then the buffer must contain _____ acid than the basemore
  13. 13what happens when you add acid to a buffer-added acid is neutralized
    - neutralizing H⁺ uses base A⁻ creating HA
    conjugate acid: H⁺ + A⁻ → HA
    - decreases pH of buffer
  14. 14what happens when you add base to a buffer-added base is neutralized
    - neutralizing OH⁻ uses conjugate acid HA creating base A⁻: OH⁻ + HA → H₂O + A⁻
    - increases the pH of the buffer
  15. 15what happens when you add base to a buffer (visual)Enter text here...what happens when you add base to a buffer (visual)
  16. 16calculating pH change in a buffer solution after adding small amt of acid/base: step 1 buffer has: HC₂H₃O₂: 0.100 mol NaC₂H₃O₂: 0.100 mol & Kₐ 1.8 × 10⁻⁵ NaOH: 0.015 mol added find pHstep 1: calculate initial []calculating pH change in a buffer solution after adding small amt of acid/base: step 1 buffer has: HC₂H₃O₂: 0.100 mol NaC₂H₃O₂: 0.100 mol...
  17. 17calculating pH change in a buffer solution after adding small amt of acid/base: step 1 explain- Eq 4 addition of NaOH to buffer
    - 1st row: init [] of acid & base
    - 2nd row: [NaOH] added, [NaOH] subtracted from [acid] and + to [base]
    - NaOH used to ionize acid/create base
    - 3rd row: new [acid], [base] & [NaOH]. [] used 4 ICE where remaining [acid] ionizes w/water
  18. 18calculating pH change in a buffer solution after adding small amt of acid/base: step 2 buffer has: HC₂H₃O₂: 0.100 mol NaC₂H₃O₂: 0.100 mol & Kₐ 1.8 × 10⁻⁵ NaOH: 0.015 mol added find pHEnter text here...calculating pH change in a buffer solution after adding small amt of acid/base: step 2 buffer has: HC₂H₃O₂: 0.100 mol NaC₂H₃O₂: 0.100 mol...
  19. 19calculating pH change in a buffer solution after adding small amt of acid/base: step 2 explain- Eq 4 remaining [acid] from step 1 as it ionizes w/water
    - 1st row: init [] of acid from step 1
    - 2nd row: unknown [acid] ionized (weak)
    - 3rd row: new [acid] after ionization
    - Ka used to determin [H₃O⁺] and pH
  20. 20calculating pH change in a buffer solution after adding small amt of acid/base: Henderson-Hasselbalch equation buffer has: HC₂H₃O₂: 0.100 mol NaC₂H₃O₂: 0.100 mol & Kₐ 1.8 × 10⁻⁵ NaOH: 0.015 mol added find pHUse ICE to get final [acid] and [base] in molescalculating pH change in a buffer solution after adding small amt of acid/base: Henderson-Hasselbalch equation buffer has: HC₂H₃O₂: 0.100...
  21. 21when calculating change in buffers containing a base and its conjugate acid, calculate pKₐ for the _____ acid of the _____ baseconjugate
    weak
  22. 22pKₐ + pKᵇ =14
  23. 23calculate pH change 4 buffer solution of a base w/conjugate acid, after adding small amt of acid/base: step 1 buffer has: NH₃ = 0.50 M NH4Cl = 0.20 M & Kᵇ = 4.75 HCl: 30 ml added find pHstep 1: find pKₐ
    pKₐ + pKᵇ = 14
    pKₐ = 14 - 4.75
    pKₐ = 9.65
  24. 24calculate pH change 4 buffer solution of a base w/conjugate acid, after adding small amt of acid/base: step 2 buffer has: NH₃ = 0.50 M NH4Cl = 0.20 M & Kᵇ = 4.75 HCl: 30 ml added find pHstep 2: get moles of added substance
    (1 M HCl / 1L) × (.300 L) = .03 M of HCl
  25. 25calculate pH change 4 buffer solution of a base w/conjugate acid, after adding small amt of acid/base: step 3 buffer has: NH₃ = 0.50 M NH₄Cl = 0.20 M & Kᵇ = 4.75 HCl: 30 ml added find pHcalculate pH change 4 buffer solution of a base w/conjugate acid, after adding small amt of acid/base: step 3 buffer has: NH₃ = 0.50 M NH...
  26. 26an effective buffer neutralizes ______ to _____ amounts of added acid or basesmall
    moderate
  27. 27factors that influence effectiveness of a buffer- relative amts of acid and conjugate base
    - absolute [] of the acid and conjugate base
  28. 28capacity of bufferhow much added acid or base it can effectively neutralize
  29. 29range of bufferthe pH range over which a particular acid and its conjugate base can be effective
  30. 30a buffer is most effective when the [] of acid and conjugate base are _____equal
  31. 31a buffer is least effective as the difference in the amts of acid and conjugate base _____increase
  32. 32guideline: effective buffer must have a [base]/[acid] ratio in the range of _____ to _____. [] of base/acid shouldn't differ by more than a factor of0.1
    10
    10
  33. 33a buffer is most effective whent [] of acid and conjugate base are _____high
  34. 34the more _____ the buffer components, the less effective the buffedilute
  35. 35the effective range for a buffering system is one pH unit on either side of _____pKₐ
  36. 36a weak acid has a pKₐ of 5.0. what is its effective range?pH range of 4.0 - 6.0
  37. 37Preparing a buffer example: which would you mix w/sodium salt to make a buffer w/pH of 4.25? HCHO₂ w/pKₐ = 3.74 or HNO₂ w/pKₐ = 3.34pH = pKₐ + log (base/acid)
    4.25 = 3.74 + log(base/acid)
    0.51 = log(base/acid)
    10^0.51 = log(base/acid)
    3.24
  38. 38buffer capacity- amt of acid/base that can be added to a buffer w/out causing a large change in pH
  39. 39buffer capacity _____ w/increasing absolute concentrations of the buffer componentsincreases
  40. 40buffer capacity _____ increases as the relative [] of the buffer components become _____ to each otherincreases
    similar
  41. 41as the ratio of the buffer components get closer to _____, the overall capacity of the buffer becomes _____one
    greater
  42. 42acid-base titration- basic/acidic solution of unknown [] reacts w/acidic/basic solution of known []
    - known solution slowly added to unknown
    - pH monitored
    - acid/base mix & neutralize each other
    - ends at equivalence point
  43. 43equivalence point- when # of mole of base = # of moles of acid initially in solution
  44. 44titration curve or pH curvea plot of the pH of the solution during a titration
  45. 45pH changes _____ near the equivalence point. small amts of added base cause _____ changes in pHquickly
    large
  46. 46Strong Acid w/Strong Base Titration 20.0 mL of 0.125 M HNO3 w/ 0.150 M NaOH step 1: get initial pH before adding basestrong acid [H₃O+] = M
    pH = -log(0.125) = 0.90
  47. 47Strong Acid w/Strong Base Titration 20.0 mL of 0.125 M HNO3 w/ 0.150 M NaOH step 2: get volume of base needed to reach equivalenceStrong Acid w/Strong Base Titration 20.0 mL of 0.125 M HNO3 w/ 0.150 M NaOH step 2: get volume of base needed to reach equivalence
  48. 48Strong Acid w/Strong Base Titration 20.0 mL of 0.125 M HNO3 w/ 0.150 M NaOH step 3: add 5 mL of base - get molesStrong Acid w/Strong Base Titration 20.0 mL of 0.125 M HNO3 w/ 0.150 M NaOH step 3: add 5 mL of base - get moles
  49. 49Strong Acid w/Strong Base Titration 20.0 mL of 0.125 M HNO3 w/ 0.150 M NaOH step 4: add 5mL of base - get final []Strong Acid w/Strong Base Titration 20.0 mL of 0.125 M HNO3 w/ 0.150 M NaOH step 4: add 5mL of base - get final []
  50. 50Strong Acid w/Strong Base Titration 20.0 mL of 0.125 M HNO3 w/ 0.150 M NaOH step 5: add 5mL of base - get [H₃O⁺]- [H₃O⁺] = (H₃O⁺ moles / ttl volume)
    [H₃O⁺] = (1.75 × 10⁻³) / (0.020 L + 0.005L)
    [H₃O⁺] = 7 × 10⁻²
    0.020 L = initial volume
    0.005L = added base volume
  51. 51Strong Acid w/Strong Base Titration 20.0 mL of 0.125 M HNO3 w/ 0.150 M NaOH step 6: add 5mL of base - pHpH = -log [H₃O⁺]
    pH = -log (7 × 10⁻²)
    pH = 1.15
  52. 52Strong Acid w/Strong Base Titration 20.0 mL of 0.125 M HNO3 w/ 0.150 M NaOH step 7: add ttl of 0.167 L of base (volume for equivalence) get pH @ equivalence point- pH 7
    - strong base has completely neutralized the strong acid
    - only source of H₃O+ is the ionization of water
  53. 53Strong Acid w/Strong Base Titration 20.0 mL of 0.125 M HNO3 w/ 0.150 M NaOH step 8: add 5 mL past equivalence point (0.0217 L) and get OH⁻ moles addedpast equivalence point there is
    [OH⁻] > [H₃O⁺]
    moles OH⁻ added =
    0.0217 L × 0.150 M NaOH = 0.0033
  54. 54Strong Acid w/Strong Base Titration 20.0 mL of 0.125 M HNO3 w/ 0.150 M NaOH step 9: add 5 mL past equivalence point (0.0217 L) and get OH⁻ moles left overOH⁻ moles left over = OH⁻ added - init H₃O⁺
    0.000755 = 0.0033 - .0025
  55. 55Strong Acid w/Strong Base Titration 20.0 mL of 0.125 M HNO3 w/ 0.150 M NaOH step 10: add 5 mL past equivalence point (0.0217 L) and get [OH⁻][OH⁻] = 0.000755 / (0.020 L + 0.0217L)
    [OH⁻] = 0.018
    0.020 L = initial volume
    0.0217L = added base volume
  56. 56Strong Acid w/Strong Base Titration 20.0 mL of 0.125 M HNO3 w/ 0.150 M NaOH step 11: add 5 mL past equivalence point (0.0217 L) and get pHpOH = -log([OH⁻])
    pOH = -log(0.018)
    pOH = 1.74
    pH = 14 - 1.74 = 12.26
  57. 57Strong Acid w/Strong Base Titration graphStrong Acid w/Strong Base Titration graph
  58. 58Strong Acid w/Strong Base Titration The initial pH is the pH of the strong _____ solutionacid
  59. 59Strong Acid w/Strong Base Titration B4 equivalence point, _____ is in excess. Calculate it by subtracting the # of moles of added _____ from the initial # of moles of _____ and dividing by _____ _____.H₃O⁺
    OH⁻
    H₃O⁺
    total volume
  60. 60Strong Acid w/Strong Base Titration @ equivalence point pH =7
  61. 61Strong Acid w/Strong Base Titration Beyond equivalence point, _____ is in excess. Calculate it by subtracting the # of moles of added _____ from the initial # of moles of _____ and dividing by _____ _____.OH⁻
    H₃O⁺
    OH⁻
    total volume
  62. 62Weak Acid w/Strong Base Titration 20.0 mL of 0.105 M CH₃COOH (Kₐ 1.8x10⁻⁵) w/ 0.125 M NaOH step 1: get initial pH before adding baseWeak Acid w/Strong Base Titration 20.0 mL of 0.105 M CH₃COOH (Kₐ 1.8x10⁻⁵) w/ 0.125 M NaOH step 1: get initial pH before adding base
  63. 63Weak Acid w/Strong Base Titration 20.0 mL of 0.105 M CH₃COOH (Kₐ 1.8x10⁻⁵) w/ 0.125 M NaOH step 2: get volume of base needed to reach equivalenceWeak Acid w/Strong Base Titration 20.0 mL of 0.105 M CH₃COOH (Kₐ 1.8x10⁻⁵) w/ 0.125 M NaOH step 2: get volume of base needed to reach equ...
  64. 64Weak Acid w/Strong Base Titration 20.0 mL of 0.105 M CH₃COOH (Kₐ 1.8x10⁻⁵) w/ 0.125 M NaOH step 3: add 5 mL of base - get molesWeak Acid w/Strong Base Titration 20.0 mL of 0.105 M CH₃COOH (Kₐ 1.8x10⁻⁵) w/ 0.125 M NaOH step 3: add 5 mL of base - get moles
  65. 65Weak Acid w/Strong Base Titration 20.0 mL of 0.105 M CH₃COOH (Kₐ 1.8x10⁻⁵) w/ 0.125 M NaOH step 4: add 5mL of base - get [] and pHWeak Acid w/Strong Base Titration 20.0 mL of 0.105 M CH₃COOH (Kₐ 1.8x10⁻⁵) w/ 0.125 M NaOH step 4: add 5mL of base - get [] and pH
  66. 66Weak Acid w/Strong Base Titration 20.0 mL of 0.105 M CH₃COOH (Kₐ 1.8x10⁻⁵) w/ 0.125 M NaOH step 4: details- after adding the base, the solution contains significant amounts of both an acid and its conjugate base
    - the solution is now a buffer
  67. 67Weak Acid w/Strong Base Titration 20.0 mL of 0.105 M CH₃COOH (Kₐ 1.8x10⁻⁵) w/ 0.125 M NaOH step 5: get pH @ 1/2 of equivalence point- amt of added base = 1/2 init [acid]:
    NaOH = 1.05 x 10⁻³ moles
    - 1/2 of init amt of acid has been converted to its conjugate base
    - results in [base] = [acid]
    - for any buffer where weak acids = conjugate acid, then pH = pKₐ = 4.74
  68. 68Weak Acid w/Strong Base Titration 20.0 mL of 0.105 M CH₃COOH (Kₐ 1.8x10⁻⁵) w/ 0.125 M NaOH step 6: get [ion] to get pH @ equivalence pointWeak Acid w/Strong Base Titration 20.0 mL of 0.105 M CH₃COOH (Kₐ 1.8x10⁻⁵) w/ 0.125 M NaOH step 6: get [ion] to get pH @ equivalence point
  69. 69Weak Acid w/Strong Base Titration 20.0 mL of 0.105 M CH₃COOH (Kₐ 1.8x10⁻⁵) w/ 0.125 M NaOH step 6: details- base has ionized all [CH₃COOH]
    - only thing left is [CH₃COO⁻] → weak base
    - the solution is no longer a buffer
    - must calculate the init [CH₃COO⁻] to get pH
  70. 70Weak Acid w/Strong Base Titration 20.0 mL of 0.105 M CH₃COOH (Kₐ 1.8x10⁻⁵) w/ 0.125 M NaOH step 7: get Kᵇ to get pH @ equivalence pointWeak Acid w/Strong Base Titration 20.0 mL of 0.105 M CH₃COOH (Kₐ 1.8x10⁻⁵) w/ 0.125 M NaOH step 7: get Kᵇ to get pH @ equivalence point
  71. 71Weak Acid w/Strong Base Titration 20.0 mL of 0.105 M CH₃COOH (Kₐ 1.8x10⁻⁵) w/ 0.125 M NaOH step 8: get pH @ equivalence pointWeak Acid w/Strong Base Titration 20.0 mL of 0.105 M CH₃COOH (Kₐ 1.8x10⁻⁵) w/ 0.125 M NaOH step 8: get pH @ equivalence point
  72. 72Weak Acid w/Strong Base Titration 20.0 mL of 0.105 M CH₃COOH (Kₐ 1.8x10⁻⁵) w/ 0.125 M NaOH step 8: details- pH at the equivalence point is not neutral but basic
    - all acid has been converted into its conjugate base, resulting in a weakly basic solution
  73. 73Weak Acid w/Strong Base Titration 20.0 mL of 0.105 M CH₃COOH (Kₐ 1.8x10⁻⁵) w/ 0.125 M NaOH step 9: get [OH⁻] 5mL after equivalence point[OH⁻] = 6.25 × 10⁻⁴ / (0.020 L + 0.0218L)
    [OH⁻] = 0.0149 M
    6.25 × 10⁻⁴ = amt of OH remaining
    0.020 L = initial volume
    0.0218L = added base volume
    - solution mix of strong base / weak base
    - strong base overwhelms weak base
    - calculate pH by strong base
  74. 74Weak Acid w/Strong Base Titration 20.0 mL of 0.105 M CH₃COOH (Kₐ 1.8x10⁻⁵) w/ 0.125 M NaOH step 10: get pH 5 mL after equivalence pointpOH = -log [OH⁻]
    pOH = -log (0.0149) = 1.83
    pH = 14 - 1.83 = 12.17
  75. 75Weak Acid w/Strong Base Titration graphWeak Acid w/Strong Base Titration graph
  76. 76Weak Acid w/Strong Base Titration initial pH is that of the _____ acid solutionweak
  77. 77Weak Acid w/Strong Base Titration between the init pH and equivalence point, the solution becomes a ____buffer
  78. 78Weak Acid w/Strong Base Titration 1/2 way to the equivalence point, the buffer components are _____ and pH =equal
    pKa
  79. 79Weak Acid w/Strong Base Titration @ equivalence point, the acid has _____ been converted into its _____ _____all
    conjugated base
  80. 80Weak Acid w/Strong Base Titration calculate the [] of ion acting as weak base by dividing the # of _____ of _____ by the ttl _____ @ the equivalence pointmoles
    ion
    volume
  81. 81Weak Acid w/Strong Base Titration beyond the equivalence point _____ is in excess. ignore the weak base and calculate the _____ by subtracting init # of moles of the ____ acid from the # of moles of added _____ and dividing by the _____ _____OH⁻
    [OH⁻]
    weak
    OH⁻
    total volume
  82. 82can monitor pH of a titration w/ a _____ _____ or an _____. With an indicator, the _____ is when it changes color and determines the _____ _____.pH meter
    indicator
    endpoint
    equivalence point
  83. 83solubility-product constant Kₛₚ- equilibrium constant for a chemical equation representing the dissolution of an ionic compound
    - has only 1 value at a given temp
  84. 84solubility of a compound is- the quantity of the compound that dissolves in a certain amount of liquid
    - can have different values in different kinds of solutions
  85. 85molar solubilitythe solubility in units of moles per liter mol/L
  86. 86Kₛₚ formulaKₛₚ formula
  87. 87relationship between Kₛₚ and molar solubility depends on the _____ of the dissociation reaction. can only compare Kₛₚ values for different compounds if the compounds have the same dissociation _____.stoichiometry; stoichiometry
    i.e. Mg(OH)₂ and CaF₂
    both have same stoichiometry; 1 mol of each produces 3 mol of dissolved ions
  88. 88solubility of an ionic compound is _____ in a solution containing a common ion than in pure water. equilibrium shifts to the _____. solubility is _____.lower
    left
    decreased
  89. 89calculate the molar solubility in the presence of a common ion: step 1 what's the molar solubility of CaF₂ in a solution containing 0.100 M NaFstep 1: write reaction eq & ksp eq
    CaF₂ ⇄ Ca²⁺ + 2F⁻
    Kₛₚ = [Ca²⁺] [F⁻]²
  90. 90calculate the molar solubility in the presence of a common ion: step 2 what's the molar solubility of CaF₂ in a solution containing 0.100 M NaFcalculate the molar solubility in the presence of a common ion: step 2 what's the molar solubility of CaF₂ in a solution containing 0.100...
  91. 91solubility of an ionic compound w/a strong or weak basic anion _____ with increasing acidity (decreasing pH)increases
    BaF₂: soluble in acidic solution because F₂ is a base
    AgI: not solubility not greater in acidic soltuion, I- is not a base
  92. 92Q (reaction quotient) and Kₛₚ- Kₛₚ value of product @ equilibrium
    - Q follows same format as Kₛₚ
    - Q value of product under any conditions
    - Q used to compare solution w/differnt [] to a solution @ equilibrium
  93. 93Q < Kₛₚ- reaction proceeds to right, to product
    - unsaturated solution
  94. 94Q = Kₛₚ- reaction @ equilibrium
    - saturated solution
  95. 95Q > Kₛₚ- reaction proceeds to left, to reactants
    - solid normally precipitates
    - unsaturated solution
    - sometimes Q can remain greater than K → supersaturated solution: unstable and forms a precipitate when disturbed
  96. 96Predicting Precipitation Reactions 0.0150M of Pb(NO₃) and 0.00350 NaBr are mixed. Does a precipitate form? step 1-list possible products & their solubility
    - soluble products won't precipitate
    NaNO₃ = soluble
    PbBr₂: Kₛₚ = 4.67 × 10⁻⁶
  97. 97Predicting Precipitation Reactions 0.0150M of Pb(NO₃) and 0.00350 NaBr are mixed. Does a precipitate form? step 2- Calculate Q and compare to Kₛₚ for product w/Ksp from step 1
    PbBr₂: Kₛₚ = 4.67 × 10⁻⁶
    Q = [Pb²⁺][Br⁻]²
    Q = (0.0150)(0.00350)² = 1.84 × 10⁻⁷
    Q < K = precipitate forms
  98. 98A solution may contain several different dissolved metal cations that can be separated by selective precipitation. What is selective precipitation?A process involving the addition of a reagent that forms a precipitate w/one of the dissolved cations but not the others
  99. 99for selective precipitation to work, Kₛₚ value must differ by at least a factor of _____10³
  100. 100Finding min required reagent [] for selective precipitation [Mg²⁺] = 0.059 M, [Ca²⁺] = 0.11M can be seperated by KOH. Kₛₚ = 2.06 × 10⁻¹³ What min [OH⁻] triggers precipitationprecipitation begins when Q = Kₛₚ; this is [] above which compound precipitates
    Q = [Mg²⁺][OH⁻]²
    2.06 × 10⁻¹³ = (0.059)[OH⁻]²
    1.9 × 10⁻⁶ = [OH⁻]
  101. 101Finding [] ions left after selective precipitation: step 1 [Mg²⁺] = 0.059 M w/Kₛₚ = 2.06 × 10⁻¹³, [Ca²⁺] = 0.11M w/Kₛₚ = 4.68 × 10⁻⁶ can be seperated by KOH. What is [Mg²⁺] when Ca²⁺ begins to precipitate.step 1: get [OH⁻] when Ca²⁺ begins to precipitate
    Q = [Ca²⁺][OH⁻]²
    2.06 × 10⁻¹³ = (0.011)[OH⁻]²
    2.06 × 10⁻² = [OH⁻]
  102. 102Finding [] ions left after selective precipitation: step 2 [Mg²⁺] = 0.059 M w/Kₛₚ = 2.06 × 10⁻¹³, [Ca²⁺] = 0.11M w/Kₛₚ = 4.68 × 10⁻⁶ can be seperated by KOH. What is [Mg²⁺] when Ca²⁺ begins to precipitate.step 1: get [Mg²⁺] that remains when Ca(OH)₂ begins 2 precipitate.
    use [OH⁻] / Q = values from step 1
    Q = [Mg²⁺][OH⁻]²
    2.06 × 10⁻¹³ = [Mg²⁺][2.06 × 10⁻²]²
    4.9 × 10⁻¹⁰ = [Mg²⁺]
  103. 103a solution of _____ [acid] shows a higher value of Ka compared to a _____ [acid]less
    more
  104. 104ph = pKa then[base] = [acid]
    ph = 7
  105. 105ph > pKa then[base] > [acid]
    ph > 7
  106. 106ph < pKa then[base] < [acid]
    ph < 7

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General Chemistry (Chem II) FlashCards sobre Chapter 17: Aqueous Ionic Equilibrium, criado por Alexandra Bozan em 29-07-2019.
Criado por
Alexandra Bozan
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