 Strength of the Acid.  K W of water.  pH scale.  pH calculation:  pH of Strong monoprotic acid/base.  pH of weak monoprotic acid/base.  pH of polyprotic acid/ base.

 HF + H 2 O H 3 O + + F - Acid Conj. Acid Conj. Base K a = [F - ] [H 3 O + ] = [F - ] [H + ] [HF] [HF]  NH 3 + H 2 O NH OH - Base Conj. Acid Conj. Base K b = [OH - ] [NH 4 + ] [NH 3 ]

Strong AcidWeak Acid Equilibrium shift To the right. Almost all the original acid is dissociated. To the left. Small extent of the acid is dissociated. Conjugate strength WeakStrong Examples HCl, H 2 SO 4, HClO 4 CH 3 COOH, H 3 PO 4, C 6 H 5 COOH

 Water is amphoteric in nature = Can act as acid and base.  Equilibrium expression of water: Kw= [H + ] [OH - ] = 1.0x at 25°C = 0.114x at 0°C = 5.47x at 50°C = 49x at 100°C Kw: Ion product constant OR dissociation constant

 There are three possible situations: A neutral solution, where [H + ] = [OH - ] An acidic solution, where [H + ] > [OH - ] A basic solution, where [H + ] < [OH - ]

 measures the hydrogen ion concentration. pH=-log[H+] pOH=-log[OH-]  Pure water has a pH very close to 7 at 25 °C. Solutions with a pH less than 7 are said to be acidic. Solutions with a pH greater than 7 are basic or alkaline.

pH is a logarithmic scale

K w = [H + ][OH - ]  -log K w = -log [H + ] – log[OH - ]  pK w = pH + pOH Since K w = 1.0x at 25°C   pH + pOH = 14.00

 Strong/weak monoprotic Acid  Strong acid: HCl H + +Cl - [H+]=[HCl] (as the ACID is almost completely dissociated) pH=-log[H + ] =-log[HCl]  Weak acid: HF H + +F - Ka= [F - ] [H + ] = [X][X] = [X] 2 [HF] [HF - X] [HF] X= [F - ]=[H + ]= √Ka[HF]

 Strong/weak monoprotic Base  Strong Base: NaOH Na + +OH - [OH - ]=[NaOH] (as the Base is almost completely dissociated) pOH=-log[OH - ] =-log[NaOH]  Weak Base: NH 4 OH NH 4 + +OH - Kb= [NH 4 + ] [OH - ] = [X][X] = [X] 2 [NH 4 OH] [NH 4 OH-X] [NH 4 OH] X= [NH4 + ] = [OH - ] = √Kb [NH 4 OH]

 Weak polyprotic acid: As observed K a1 >K a2- Therefore, [H + ]= √K a1 [H 2 CO 3 ]

 It is useful to specify the amount of weak acid/base that has dissociated in achieving equilibrium. %Dissociation=amount dissociated X 100 initial concentration e.g. % Dissociation = [H + ] X 100 [HF]

 Question 1: The pOH of a sample of backing soda dissolved in water is 5.74 at 25°C. Calculate the pH, [H + ] and [OH - ] for this sample. Is the solution acidic or basic?

 Question 1 answer: pKw=pH + pOH=14 ⁂ pH= =8.26 ⁂ [H + ]=Shiftlog (-8.26)=5.49x10 -9 M ⁂ [OH - ]=1.0X / 5.49x10 -9 M= 1.82X10 -6 M The solution is basic

 Question 2 Calculate the pH in each of the following: a)0.005 M HNO 3 b) 0.6 M KOH c) 0.25 M CH 3 COOH (Ka = 1.8x10 -5 ) d) 2M NH 3 (Kb= 1.8x10 -5 ) e) 0.2M Arsenic acid (H 3 AsO 4 ) ( Ka 1 = 5 x 10 -3,Ka 2 = 8x10 -8,Ka 3 = 6x )

 Question 2 answer a) [H + ]=[HNO 3 ]=0.005M ⁂ pH= 2.30 b) [OH - ]=[KOH]=0.6M ⁂ pOH= 0.22 ⁂ pH= c)CH 3 COOH CH 3 COO - + H + Ka= [CH 3 COO - ] [H + ]/ [CH 3 COOH] ⁂[H + ]= √Ka [CH 3 COOH] = √(1.8x )(0.25)=2.12x10 -3 M ⁂pH=-log(2.12x10 -3 )=2.67

 Question 2 answer (Cont.) d) NH 3 +H 2 O NH 4 + +OH- Kb=[NH 4 + ][OH-]/[NH 3 ] ⁂[OH-]= √KB [NH 3 ] = √(1.8x10 -5 )(2)=6. 0x10 -3 M ⁂pOH=-log(2.12x10 -3 )=2.22 ⁂pH=11.78

 Question 2 answer (Cont.) e) H 3 AsO 4 H + + H 2 AsO 4 - Ka 1 = 5 x H 2 AsO 4 - H + + HAsO 4 2- Ka 2 = 8x10 -8 HAsO 4 2- H + + AsO 4 3- Ka 3 = 6x Ka=[H + ][H 2 AsO 4 - ]/[H 3 AsO 4 ] ⁂[H + ]=√Ka [ H 3 AsO 4 ]=√(5x10 -3 )(0.2)=0.03 M ⁂pH=-log(0.03)=1.50

 Question 3 Determine the percent dissociation of formic acid HCOOH in a 2M solution. (Ka= 1.77 x10 -4 )

 Question 3 answer: HCOOH H + + HCOO - Ka=[H + ] [HCOO-]/ [HCOOH] ⁂[H + ]= √Ka [ HCOOH ] = √(1.77x10 -4 )(2) = 0.018M ⁂% Dissociation=(0.018/2)x100=0.94%

 Question 4: Given the K a value for acetic acid is 1.8 x and the K a value for hypochlorous (HOCl) acid is 3.5 x Which is the stronger base, OCl - or CH 3 COO - ? Since Ka acetic acid > Ka HOCl ⁂ CH 3 COO - < OCl -

 Question 4 The pH of a sample of human blood was measured to be 7.41 at 25°C. Calculate pOH, [H + ] and [OH - ] for the sample.  pH + pOH =   pOH = = 6.59 [H + ] = antilog(-pH)= antilog (-7.41) = 3.9x10 -8 M [OH - ] = antilog(-pOH) = antilog (-6.59) = 2.6x10 -7 M 24

1- In a 1.00M solution of HF, the percent dissociation is 8.1%. Calculate Ka. [6.5 x ] 2- Calculate the concentration of an aqueous KOH solution that has pH= [3.16 x M] 3- Arsenic acid (H 3 AsO 4 ) is a triprotic acid with ( Ka 1 = 5 x 10 -3,Ka 2 =8x10 -8,Ka 3 = 6x ). Calculate the concentration of [HAsO 4 2- ] in a 0.20 M arsenic acid. [ 8x10 -8 M ]