http://lawrencekok.blogspot.com

Prepared by Lawrence Kok

Video Tutorial on Acids Bases, pH, Acid Dissociation Constant Ka, Ionic Product Water Kw, pKa and pKb

Acids and 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 gas2HCI +  Mg → MgCI2 + H2

• Acid with Ammonia (aq) - Salt + WaterHCI + 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  + WaterCa(OH)2 + HCI → CaCI2 + H2O

Metal Oxide -  Salt and WaterCaO + 2HCI → CaCI2 + H2O

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 from 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 NaOH1st Method using pOH[OH] = 0.1MpOH = -log[OH] = -log [0.1] = 1pH + pOH = 14pH + 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]

• Kw  = 1.0x10-14 mol2 dm-6

Kw is Temp dependent, dissociation water to ions is endothermic. 

H2O + H2O ↔ H3O+ + OH −    ΔH = +57kJ/molKw = [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 and pH is 7 (neutral)

• 50C, Kw is 9.3 x 10-14 and pH is 6.5 (neutral)

• 50C Higher Temp, pH 6.5 (neutral) < 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

Base Dissociation Constant, Kb

Ka calculation. Click HERE to viewKb calculation. Click HERE to view

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 pKaWeak acid - Low Ka, High pKaAcidicty 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 2.30) more acidic than Ethanoic acid (pH 2.88) cause has Higher Ka/Low pKa

Click HERE to view Ka for different acidsClick HERE to view more info on acids and bases

Strong vs Weak 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)           (0.01mole 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

Concentrated Vs Diluted

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 FormulaDissociation Acid, HA = KaHA + H2O ↔H3O+ + A-Ka = [H3O+][A- ]/ [HA]

Dissociation Base, A- = KbA- + H2O ↔ HA + OH-

Kb = [HA][OH-] /[A-]Ka x Kb = Kw (see diagram)

3rd FormulaKa x Kb = Kw(-log both sides)

- log { Ka x Kb } = - log { Kw } - log Ka - log Kb = - log KwpKa + 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]         • [CH3COOH] = [H+] [CH3COO-] / Ka

• pH = -log [H+], 4.50 = -log [H+], [H+] = 3.16 x 10-5

• [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 = 10.6

Calculate pKb for 0.03M NH3 given pH 10.0 NH3 + H2O ↔ NH4

+ + OH-

Kb = [NH4+][OH-] / [NH3]

• Kb = [OH-]2/ [NH3] • pH + pOH = 14, pOH = 14 - 10 = 4, pOH = - log [OH],  4 = -log [OH], [OH] = 10 -4

• 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

CH3NH2 + H2O ↔ CH3NH3+  + OH-Kb = [CH3NH3+][OH- ] / [CH3NH2 ]

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

• 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 approximate calculationClick HERE to view quadratic calculation

Acknowledgements

Thanks to source of pictures and video used in this presentation

Thanks to Creative Commons for excellent contribution on licenseshttp://creativecommons.org/licenses/

Prepared by Lawrence Kok

Check out more video tutorials from my site and hope you enjoy this tutorialhttp://lawrencekok.blogspot.com