Question

Fill in the blanks and please show all work for Calculated Ka or Kb column and explain how to find which ion is hydrolyzed ?

Part 1: pH of Solutions of Salts: In this part you will measure the pH values for 0.1 M solutions of the following solutes. Sodium bicarbonate o M Copper(II) sulfate oOM Iron(III) nitrate 010M Sodium acetate O-10M Ammonium chloride O Ib M Sodium fluoride o10 M Methylammonium chloride o.o M For each solution, place the solution into a small beaker. Rinse and blot the pH electrode, place the pH electrode into the solution, and read the pH by moving the knob on the console to pH. Then return the knob to Standby before moving the electrode out of the solution for rinsing. Record the pH for each solution determine the ion that influences the pH and the equilibrium involved. Finally calculate the equilibrium constant based on on your data sheet and then the pH value you measured. Hydrolysis and Buffers Laboratory Report Name: Date: Partner's Name: Instructor's Initials: Part 1: pH of Solutions of Salts: Which ion Calculated K, or Kb Solution Measured pH hydrolyzed? NaC;H,O 779 NH,CI 4 83 Sh d Basic NaF 4-55 ea NAHCO 9.00 CHINH CI S.94 Acidic pidic CUSO 3.90 Fe(NOs)s t34 SHOW YOUR CALCULATIONS FOR K, AND K ON A SEPARATE SHEET. Hydrolysis Reaction (write the net ionic equation) Solution NaC;H3O NH CI NaF NaHCO CHINH&CI CUSO Fe(NO.)s Acid E Acaie

Answer 1

1) Sodium acetate (NaCH₃COO) : 0.1 M

NaCH₃COO -----> Na⁺ + CH₃COO^-

CH₃COO^- + H₂O -----> CH₃COOH + OH^-

Initial 0.1    0 0

   Change -x +x +x

   equilibrium 0.1-x x    x

Kb = [CH₃COOH] [OH^-]/[ CH₃COO^-]

pH = 7.79 , pOH = 14-7.79 = 6.21, [OH^-] = 10^-pOH = 10^-6.21 = 6.2 x 10^-7 M

[OH^-] = [ CH₃COOH] = 6.2 x 10^-7 M

As 6.2 x 10^-7 M is very less number it can be neglected in 0.1-x

Kb = (6.2 x 10^-7)²/0.1 = 38.44 x 10^-13

Ka = Kw/Ka = 10^-14/38.44 x 10^-13 = 0.00260 = 2.6 x 10^-3

2) NH₄Cl - 0.1 M :

  NH₄Cl -------> NH₄⁺ + Cl^-

  NH₄⁺ + H₂O ------> NH₃ + H₃O⁺

Initial    0.1    0 0

Change    -x +x    +x

Equilibrium    0.1-x    x    x

Ka = [H₃O⁺] [NH₃]/NH₄⁺ = x²/0.1-x

pH = 4.83 , [H₃O⁺] = 10^-pH = 10^-4.83 = 0.000015

As 0.000015 << 1 Hence it is neglected in 0.1-x

Ka = (0.000015)²/0.1 = 2.25 x 10^-9

Kb = Kw /Ka = 10^-14/2.25 x 10^-9 = 4.44 x 10^-6

3) NaF - 0.1 M

NaF ---> Na⁺ + F^-

   F^- + H₂O ---> HF + OH^-

Initial 0.1 0 0

Change    -x    +x +x

Equilibrium    0.1-x x x

Kb = [OH^- ] [HF]/[F^-]

pH = 7.55 , pOH = 14-7.55 = 6.25 , [OH^-] = 10^-pOH = 10^-6.25 = 5.6 x 10^-7 M

As 5.6 x 10^-7 M << 0.1 it is neglected in 0.1-x

Kb = (5.6 x 10^-7 )²/0.1 = 31.36 x 10^-13

Ka = Kw/Kb = 10^-14/31.36 x 10^-13 = 0.032 x 10^-1 = 3.2 x 10^-3

D) CH₃NH₃Cl -0.1 M

  CH₃NH₃Cl ----> CH₃NH₃⁺ + Cl^-

  CH₃NH₃⁺ + H₂O ---> CH₃NH₃OH + H⁺

Initial    0.1    0 0

Change    -x +x +x

Equilibrium    0.1-x    x    x

Ka = [H⁺] [CH₃NH₃OH]/[CH₃NH₃⁺ ]

pH = 5.94 , [H⁺] = 10^-pH = 10^-5.94 = 1.1 x 10^-6 M

As 1.1 x 10^-6 M << 0.1 M it is neglected in 0.1-x

Ka = (1.1 x 10^-6)²/0.1 = 1.21 x 10^-11

Kb = Kw/Ka = 10^-14/1.21 x 10^-11 = 0.826 x 10^-3 = 8.26 x 10^-4

e) CuSO₄ - 0.1 M

CuSO₄ ---> Cu⁺² + SO₄^-2

   Cu⁺² + 4H₂O --->[Cu(H₂O)₄]⁺²

[Cu(H₂O)₄]⁺² + H₂O ---> [Cu(H₂O)₃OH]⁺ + H₃O⁺

Initial    0.1 0    0

Change 0.1-x    +x    +x

Equilibrium 0.1-x    x x

Ka = [H₃O⁺] [[Cu(H₂O)₃OH]⁺ /[Cu(H₂O)₄]⁺² = X²/0.1-X

pH = 3.90, [H₃O⁺] = 10^-pH = 10^-3.90 = 1.3 X 10^-4

As 1.3 X 10^-4 << 0.1 so it is neglected in 0.1-x

Ka = (1.3 x 10^-4)²/0.1 = 1.69 x 10^-7

Kb = Kw/Ka = 10^-14/1.69 x 10^-7 = 0.592 x 10^-7 = 5.92 x 10^-8

f) Fe(NO₃)₃ - 0.1 M

  Fe(NO₃)₃ ----> Fe⁺³ + 3NO₃^-

  Fe⁺³ + 6H₂O ---> [Fe(H₂O)₆]⁺²

[Fe(H₂O)₆]⁺² --> [Fe(H₂O)₅OH⁺² + H₃O⁺

Initial    0.1 0    0

Change    -x +x +x

Equilibrium 0.1-x    x x

Ka = [H₃O⁺] [[Fe(H₂O)₅OH]⁺² /[[Fe(H₂O)₆]⁺²

pH = 1.34, [H₃O⁺] = 10^-pH = 10^-1.34 = 0.0457

Ka = (0.0457)²/(0.1-0.0457) = 0.002088/0.9543 = 0.00219

Kb = Kw/Ka = 10^-14/0.00219 = 0.00457 x 10^-9 = 4.57 x 10^-12