Tutorial 2

Answer of last week’s extra question Arsenic acid (H3AsO4) is a triprotic acid with (Ka1 = 5 x 10-3 , Ka2 =8x10-8, Ka3 = 6x10-10). Calculate the concentration of [HAsO42-] in a 0.20 M arsenic acid. H3AsO4 H2AsO4 - +H+ Ka1= 5 x 10-3 M H2AsO4 - HAsO42- +H+ Ka2=8x10-8 M Ka1 = [H2AsO4 - ] [H+ ] [ H3AsO4 ] [H2AsO4 - ] = [H+ ] = √ Ka1* [ H3AsO4 ]=0.032 M Ka2 = [HAsO4 2- ] [H+ ] H2AsO4 - Ka2 = [HAsO4 2- ] [0.032 ] [0.032] So, Ka2= [HAsO4 2- ] = 8x10-8 M

Objectives: Determine acid-base properties of salts. Calculate the pH of different salt solutions.

What is a salt? A salt is an ionic compound that dissociates into its ions once it is dissolved in water. A salt solution can be either: 1- acidic 2- basic 3- neutral

Neutral salts: Example: NaCl NaCl Na+ + Cl- Both Na+ and Cl- are weak conjugate acid and base, they don’t release or attract protons, that is to say they have no effect on pH of water. H2O Salts that consist of : Cations of strong bases and the anions of strong acids have no effect on pH when dissolved in water. pH of the salt solution=7.00 Other examples: KCl, KNO3 and NaNO3

Basic salts: Example: KF KF K+ + F- F- +H2O HF +OH- How to calculate its pH ? Kb= [HF] [OH-] = [x] [x] [ F-] [ F-] [OH-]= √ Kb[F-] p[OH]=-log √ Kb[F-] pH= 14 - pOH H2O Weak Conj. acid Strong Conj. base [Solution is Basic] NOTE: Kw= Ka x Kb where Ka is the dissociation constant of weak acid and Kb is the dissociation constant of its conj. Base.

Basic salts (Cont.): Other examples: Salts that consists of: Cations of strong bases and anions of weak acids are basic salts, when they are dissolved in water, pH>7.00 Other examples: CH3COOK, Na2CO3 and NaHCO3

Acidic salts Example: NH4Cl NH4Cl NH4+ + Cl- NH4+ + H2O NH3 + H3O+ Strong Conj. acid Weak Conj. base NH4+ + H2O NH3 + H3O+ How to calculate its pH ? NOTE: Kw= Ka x Kb where Ka is the dissociation constant of weak acid and Kb is the dissociation constant of its conj. Base. Ka= [ NH3] [ H+] [NH4+] = [x] [x] [NH4+] [ H+]=√Ka [NH4+] pH= -log √ Ka [NH4+]

Acidic salts (Cont.) Another type of acidic salts: Salts that consist of: anions of strong acids and cations of weak bases are acidic salts, when they are dissolved in water, they affect the pH of water where it becomes less than 7.00 Another type of acidic salts: Salts that contain a highly charged metal ion. Ex: AlCl3 These salts form hydrated complexes when dissolved in H2O that react as acids. pH= - log √ Ka[AlCl3]

Salt whose cation and anion come from a weak base and a weak acid: To predict whether they form neutral, acidic or basic solutions, you have to compare the Ka of the weak acid with the Kb of the weak base. If: Ka = Kb so neutral salt pH = 7.00 Ka > Kb so acidic salt pH less than 7.00 Ka < Kb so basic salt pH greater than 7.00 Example: CH3COONH4 ka CH3COOH = 1.8 X 10-5 Kb NH4OH = 1.8 X 10-5 Ka = Kb so the salt is neutral.

Question 1: Arrange the following 0.1 M solution in order of most acidic to most basic. KOH, KCl, KCN, NH4Cl and HCl. HCl > NH4Cl > KCl > KCN > KOH

Question 2: Predict whether a 0.1 M aqueous solution of each of the following salts will be acidic, basic or neutral and calculate its pH a) NH4C2H3O2 b) NH4CN c) NH4F (Ka of CH3COOH = 1.8X10-5, Kb of NH3 = 1.8X10-5, Ka of HCN = 6.2X10-10 , Ka of HF = 7.1x10-4) a) NH4C2H3O2 Ka of CH3COOH = Kb of NH3 So the salt is neutral, pH= 7.00

pOH = - log √ Kb[CN-] =2.89 pH=11.10 b) NH4CN Ka of HCN < Kb of NH3 so the salt is basic, pH>7.00 To calculate its pH: NH4CN NH4+ + CN- CN- + H2O HCN + OH- Kb= [HCN] [OH-] [CN-] pOH = - log √ Kb[CN-] =2.89 pH=11.10 Note: Conj.base is stronger than conj. acid NOTE: Kb=Kw/Ka where Ka is the dissociation constant of weak acid and Kb is the dissociation constant of its conj. Base.

so salt is acidic , pH < 7.00 To calculate its pH: NH4F NH4+ + F- C) NH4F Ka of HF > Kb of NH3 so salt is acidic , pH < 7.00 To calculate its pH: NH4F NH4+ + F- NH4+ + H2O NH3 + H3O+ Ka=[ NH3] [ H+] [NH4+] Ka= [x][x] H+= √ Ka [NH4+] pH= -log √ Ka [NH4+] pH= 5.13 Note: Conj. Acid is stronger than conj. base Kb=Kw/Ka 1x10-14/ 1.8x10-5 = 5.55x10-10

Question 3: An unknown salt is NaCN, KNO3, NaF, NaCl, or NH4Cl. When 0.1 mol of the salt is dissolved in 1.0 L of solutions, the pH of the solution is 8.07. What is the identity of the salt? Ka of HCN = 6.2 x l0-10 Ka of HF = 7.2 x10-4 It is either NaF or NaCN as they are the only basic salts present. So we have to calculate their pH. For NaF pOH= -log √ Kb[NaF] pOH= -log √ 1.38x10-11x0.1= 5.93 pH= 14-5.93= 8.07 If you calculate the pH of NaCN it is equal 11.10 So the unknown salt is NaF. Kb=Kw/Ka = 1x10-14 / 7.2x10-4 = 1.38x10-11

Question 4: Calculate the equilibrium concentration of HCN in a 0.05 M NaCN solution ( Ka for HCN=6.2x10-10). NaCN Na+ + CN- 0.05M 0.05M 0.05M CN- + H2O HCN + OH- Kb= [x][x] [CN-] X2= Kb [CN-] X2= 1.61x10-5[0.05] X= HCN= 8.9x10-4 M H2O Kb=Kw/Ka = 1x10-14 / 6.2x10-10 = 1.61x10-5

Question 5: Kb= [x] [x] [N3- ] X2 =5.26x10-10 [0.01] Sodium azide (NaN3) is sometimes added to water to kill bacteria. Calculate the concentration of all species in a 0.01 M solution of NaN3. The Ka value for hydrazoic acid (HN3) is 1.9 x 10-5 NaN3 Na+ + N3- 0.01M 0.01M N3- + H2O HN3 + OH- X X Kb= [x] [x] [N3- ] X2 = Kb [N3- ] X2 =5.26x10-10 [0.01] X= [HN3 ]=[OH- ]= 2.26x10-6 M H2O The species are : Na+, N3-, HN3, OH-

Question 6: Calculate the pH of a 0.25M solution NH4NO3 Kb of NH3= 1.8 X10-5. NH4NO3 NH4+ + NO3- NH4+ + H2O NH3 + H3O+ acidic solution Ka= [NH3] [H+] [NH4+] Ka= [x][x] X2= Ka [NH4+] X2= 5.55x10-10 [0.25] X=H+=1.17X10-5 M pH= -log[1.17X10-5]=4.92 Kb=Kw/Ka = 1x10-14 / 1.8x10-5 = 5.55x10-10

Question 7: Calculate the pH of 0.1 M Na3PO4. ka3 = 4.8x10-13 Na3PO4 3Na+ + PO43- PO43- + H2O HPO42- + OH- Kb= [HPO42- ][OH- ] [PO43-] Kb = x2 0.1 X=OH- = 0.0447M pOH = 1.35 pH= 12.65 Kb=Kw/Ka3 = 0.02

Question 8: Calculate the pH of 0.01 M AlCl3 solution. The Ka value of Al(H2O)63+ = 1.4X10-5 AlCl3 Al3+ + Cl- Al3+ + 6H2O Al(H2O)63+ Al(H2O)63+ + H2O [Al(H2O)5OH]2+ + H3O+ Ka = [Al(H2O)5OH]2+ [H+] Al(H2O)63+ [H+]2 = Ka [Al(H2O)63+]= 1.4x10-7 [H+] = 3.74x10-4 M pH = 3.43

Extra problems: Calculate the pH of a 0.20 M K2S solution (Ka2 for H2S= 1 x 10-13). pH=13.16 Calculate the pH of a 0.45 M solution of sodium acetate, CH3COONa, (Ka of CH3COOH = 1.8X10-5) pH=9.19 Calculate the pH of a 0.25M solution of NaCl. pH=7.00 Calculate the pH of 0.4 M solution NH4NO2, ( Ka of HNO2 = 4.5 X10-4 and Kb of NH3= 1.8x10-5). pH=4.82