Monday, July 18, 2011

IB Chemistry Acids Bases, pH, Acid Dissociation Constant Ka, Ionic Product Water Kw, pKa, pKb

Acids Bases, pH scale, Acid Dissociation Constant Ka, Water Dissociation Constant Kw, pKa 


Physical properties of Acids
  • Electrolytes, produces H3O+, hydronium ion, in water (conduct electricity)
  • Sour, pH <7, turns litmus red, phenolphthalein colourless, methyl orange to red
    Physical properties of Bases
    • Bitter, pH> 7, turns litmus blue, phenolphthalein pink, methyl orange to yellow
    Chemical properties of Acids / Bases
    • Acid with Metal - Salt + H2 gas
    • 2HCI +  Mg  ---->  MgCI2 + H2
    • Acid with Ammonia (aq) - Salt + Water
    • HCI + NH4OH -----> NH4CI + H2O
    • Acid with Bases (Neutralization) - Salt + Water and other products. Bases are - Metal Hydroxide, Metal Oxide, Metal Carbonates
    Acid HCI reacts with Bases :
    • Metal Hydroxide (Soluble or Alkali ) -  Salt + Water 
    • LiOH + HCI---> LiCI + H2O
    • Metal Hydroxide (Insoluble) -  Salt  + Water
    • Ca(OH)2 + HCI ----> CaCI2 + H2O
    • Metal Oxide -  Salt and Water
    • CaO + 2HCI ---> CaCI2 + H2O




    • Metal Carbonate -  Salt, Water and Carbon dioxide
    • CaCO3 + 2HCI ---> CaCI2 + H2O + CO2
    .......................................................................................................................................................
    pH measurement of Acidity of solution
    pH scale for Acid
    • pH is logarithmic scale
    • Relationship between pH with Conc H+
    • pH = - log [H+], p stand for - log
    • pH = Power of Hydrogen, pH 7 = Conc H+ is 10 -7
    • Conc H+= 0.0000001M, pH = -log[0.0000001], pH = 7
    • Easier using pH scale than Conc [H+] 
    • Conc H+ up 10x from 0.0001(10 -4 ) to 0.001(10 -3), pH change by 1 unit - pH 4 to 3 
    • pH 3 is (10x) more acidic than pH 4
    • 1 unit change in pH is 10 fold change in H+ Conc


    pOH for Base
    • Relationship between pOH with Conc OH
    • pOH = -log [OH]
    • [OH] = 0.0000001M, pOH = -log[0.0000001], pOH = 7
    • pH + pOH = 14, pH + 7 = 14, pH = 7
    Calculate pH for 0.1M NaOH
    1st Method using pOH
    • [OH] = 0.1M
    • pOH = -log[OH] = -log [0.1] = 1
    • pH + pOH = 14
    • pH + 1 = 14, pH = 13.0
    2nd Method using Kw
    • [H+] x [OH] = Kw
    • [H+] x [0.1] = 1 x 10 -14
    • [H+] = 1 x 10 -14/ [0.1]
    • [H+] = 10 -13
    • pH = -log[H+], pH = -log [10 -13], pH = 13.0
    ..........................................................................................................................................................
    Dissociation Constant or Ionic Product Water, Kw
    H2O dissociate forming H3O+ and OH
    H2O + H2O  <----> H3O+ + OH −
    Kw =  [H3O+][OH −]/[H2O]2
    • Dissociation H2O is small, so conc [H2O] remains constant giving Kw = [H3O+][OH −]
    • [H3O+]= 1.0x10-7, [OH] = 1.0x10-7
    • Kw = [H3O+][OH −]
    • Kw = [1.0x10-7][1.0x10-7] = 1.0x10-14mol2 dm-6
    Kw is temperature dependent, dissociation water to ions is endothermic. 
    H2O + H2O  <----> H3O+ + OH −    ΔH = +57kJ/mol
    Kw = [H3O+][OH −]
    Temp increases, equilibrium shift to right (endo), so more ions form, Kw increases
    • Kw is Temp dependent
    • Kw at 25C is 1.0 x 10-14
    • Temp up, Kw up, H+up, pH down
    • 25C, Kw is 1.0 x 10-14, pH is 7 (neutral)
    • 50C, Kw is 9.3 x 10-14, pH is 6.5 (neutral)
    • 50C Higher Temp, neutral pH is 6.5, lower than 7
    Working below.
    [    H+][OH −    ]=Kw, [H+][OH −] = 9.3 x 10-14, [H+]2= 9.3 x 10-14, [H+]     = √9.3 x 10-14
    [H+] = 3.05 x 10-7
    pH 6.5 is NEUTRAL  cause number of [H+] and [OH −] are the same = 3.05 x 10-7
    .......................................................................................................................................................
    Acid Dissociation Constant, Ka (Ka measure the strength of acid)
    HA <--> H+ + A-
    • Ka = (H+)(A-) / HA
    • Ka High - Stronger Acid
    • pKa = -log (Ka)
    • pKa Low - Stronger Acid
    Ka calculation. Click HERE to view
    Kb calculation. Click HERE to view


    Base Dissociation Constant, Kb 
    Base (A-) dissolves in water produce OH −
    A- + H2O  <----> HA + OH −
    • Kb = [HA][OH] / [A-]
    • Kb High - Stronger Base
    • pKb = - log (Kb)
    • pKb Low - Stronger Base

    Acids with High Ka or Low pKa is more acidic
    • Strong acid -High Ka, Low pKa
    • Weak acid - Low Ka, High pKa

    Acidicty 0.10M methanoic > 0.10M ethanoic 
    • Methanoic acid, Ka = 1.77 x 10-4
    HCOOH  <------> HCOO-  + H+
    [HCOO-][H+] / [HCOOH] = Ka
    [H+]2 / [HCOOH] = 1.77 x 10-4
    [H+] = √1.77 x 10-4 x 0.10 
    [H+] = 4.21 x 10-3
    pH = -log[H+] = - log [4.21 x 10-3],  pH = 2.30
    • Ethanoic acid, Ka = 1.78 x 10-5 
    CH3COOH <---->  H+ + CH3COO-
    [H+][CH3COO-] / [CH3COOH] = Ka
    [H+]2 / [CH3COOH] = 1.8 x 10-5 
    [H+] = 1.33 x 10-3
    pH = -log [H+] = -log [1.33 x 10-3], pH = 2.88
    Methanoic (pH 2.3) more acidic than Ethanoic acid (pH 2.88) cause has higher Ka/Low pKa


    Click HERE to view Ka for different acids
    Click HERE to view more info on acids and bases
    .........................................................................................................................................................
    Strong/Weak vs Dilute/Concentrated
    Strong Acid/Base - Ionises/ Dissociate completely (100%), Strong electrolyte, Higher conductivity
    HCI ------>        H+      +     CI- 
    (1 mole)        (1 mole H+)

    NaOH ----->      Na+    +    OH −
    (1 mole)                        (1 mole OH −)

    Ba(OH)2 ------>    Ba2+  +   2OH −
    (1 mole)                       (2 mole OH −)
    • Strong acids - HI, HBr, HCI, HNO3, H2SO4, HCIO3, HCIO4
    • Strong Bases - LiOH, KOH, NaOH, CsOH, Ca(OH)2, Ba(OH)2
    Weak acid - Partially dissociate, Most exist in molecules, Poor electrolyte, Lower conductivity
    CH3COOH <----->     H+     +      CH3COO-
    (1 mole)               (0..01 mole H+)
    • Weak Acid  - CH3COOH, HF, HCN, H2CO3, H3BO3, HNO2, H3PO4
    • Weak Bases  - NH3, C2H5NH2 , CH3NH2, (CH3)2NH, C3H5O2NH2
    Concentrated Acid/Base - high number/moles of solutes per dm3 solution
    Diluted Acid/Bases - low number/moles of solutes per dm3 solution
    10M CH3COOH vs 0.01M HCI
    • 10M CH3COOH is weak but concentrated acid caused not all 10M acid molecule will dissociate fully forming H+ ions
    • 0.01M HCI is strong but diluted acid cause all 0.01M acid, fully dissociate forming H+ ions
    .........................................................................................................................................................
    Important Formula for IB Calculations
    1st Formula
    • [H+][OH ] = 1.0 x10-14  (- log both sides)
    • -{ log[H+] + log [OH] } = - { log 1 + log 10-14  }    pH = - log[H+], pOH = - log [OH]
    • pH + pOH = pKw or  pH + pOH = 14
    2nd Formula
    • Dissociation Acid,HA=Ka
    • Dissociation Base, A- =Kb
    • HA + H2O <-> H3O+ + A-
    • Ka = [H3O+][A- ]/ [HA]
    • A- + H2O <-> HA + OH-
    • Kb = [HA][OH-] /[A-]
    • Ka x Kb = Kw (see diagram)


    3rd Formula
    Ka x Kb = Kw(-log both sides)
    • - log { Ka x Kb } = - log { Kw } 
    • - log Ka - log Kb = - log Kw
    • pKa + pKb = pKw or pKa + pKb = 14
    ........................................................................................................................................................
    Calculation of pH, Ka and Conc of Weak acid 
    • Calculate pH of 0.10M CH3COOH, given Ka = 1.8 x 10-5 
    CH3COOH <----->  H+ + CH3COO-
    [H+] [CH3COO-] / [CH3COOH] = Ka
    [H+]2 / 0.10 = 1.8 x 10-5 
    [H+] = 1.33 x 10-3
    pH = -log [H+] = -log [1.33 x 10-3]
    pH = 2.88
    • Calculate Ka of 0.020M CH3COOH given pH is 3.90
    CH3COOH <----->  H+ + CH3COO-
    Ka = [H+] [CH3COO-] / [CH3COOH],          pH = -log [H+], 3.90 = -log [H+], [H+] = 1.26 x 10-4
    Ka = [1.26 x 10-4] [1.26 x 10-4] / 0.020 
    Ka = 7.92 x 10-7
    • Calculate Conc of CH3COOH given Ka is 4.1 x 10-6 and pH is 4.50
    CH3COOH <----->  H+ + CH3COO-
    Ka = [H+] [CH3COO-] / [CH3COOH]         pH = -log [H+], 4.50 = -log [H+], [H+] = 3.16 x 10-5
    [CH3COOH] = [H+] [CH3COO-] / Ka
    [CH3COOH] = [3.16 x 10-5] [3.16 x 10-5]  / 4.1 x 10-6
    [CH3COOH] = 2.44 x 10-4

    Calculation of pH, Kb and Conc of Weak base
    • Calculate pH of 0.01M NH3 given Kb is 1.8 x 10-5
    NH3 + H2O <-----> NH4+ + OH-
    [NH4+][OH-] / [NH3] = Kb
    [OH-]2 / [NH3 ] = 1.8 x 10-5
    [OH-] = √1.8 x 10-7, [OH-] = 4.24 x 10-4    pOH = -log [OH], pOH = -log [4.24 x 10-4] = 3.37
    pH + pOH = 14
    pH = 14 - 3.37 = 10.6
    pH = 10.60
    • Calculate pKb for 0.03M NH3 given pH 10.0 
    pH + pOH = 14, pOH = 14 - 10 = 4, pOH = - log [OH],  4 = -log [OH], [OH] = 10-4
    NH3 + H2O <-----> NH4+ + OH-
    Kb = [NH4+][OH-] / [NH3]
    Kb = [OH-]2/ [NH3] 
    Kb = [10-4]2 / 0.03 = 3.33 x 10-7
    pKb = -log Kb = -log 3.33 x 10-7 = 6.48
    pKb = 6.48

    • Calculate the Conc base [CH3NH2 ], given pH is 10.8,  pKa is 10.64
    Convert pH/Ka/pKa to pOH/Kb/pKb cause it is a base
    pKa + pKb = 14, pKb = 14 - 10.64, pKb = 3.36
    pH + pOH = 14, pOH = 14 - 10.8, pOH = 3.2


    CH3NH2 + H2O <--> CH3NH3+  + OH-
    Kb = [CH3NH3+][OH- ] / [CH3NH2 ]
    Kb = [OH-]2 / [CH3NH2 ]
    [CH3NH2 ] = [OH-]2 / [Kb ] =  [10-3.2]2 / 10-3.36
    [CH3NH2 ] = 9.13 x 10-4
    ........................................................................................................................................................
    Video on Ka calculation using Approximation Method

    Key notes from video:
    For pH Cal and Significant figures
    • pH = -log [H+] = -log [1.3 x 10-3] = 2.88
    • Conc = 1.3 x 10-3(2 sig fig)
    • pH = 2.88 (3 sig fig)
    • Number sig fig diff for log calculation
    • IB Cal, approximate method is used instead of using quadratic

    Click HERE to view quadratic calculation
    .......................................................................................................................................................

    IB Chemistry Acids Bases, pH, Acid Dissociation Constant Ka, Ionic Product Water Kw, pKa, pKb  
    Thanks to all pictures and videos contributors for above post
    Posted by Lawrence Kok at 2:14 AM

    1 comment:

    1. This is indeed another good source of information about dissociation constant, which is similar to this one: http://dissociationconstant.com/ . You must be brilliant with this topic. Hope to learn more from you!

      ReplyDelete

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