Only part c and part d please show work I only need Part C and Part D data blank

Question

Only part c and part d please show work





I only need Part C and Part D data blanks to be finished EXPERIMENT 25BUFERS BUFFERS Objectives 1. To gain experience using a p 2. To learn about buffers meter. To be able to calculate the pl of a buffer solution addition of an acid or base. 3. solution and predict how it will react to 25.1 INTRODUCTION very slightly ionized (Git would approach 99 Even in quite di percent ionization only as the concentration approaches 0.0M) quite dilute solutions, acetic acid is s, acetic acid is On the other hand, the salt, sodium acetate trihydrate (NaC,H,o, 3H,O), is 100 percent dissociated into the constituent hydrated ions in a dilute aqueous Sodium acetate solutions are basic because the acetate ion (the conjugate base of acetic acid) behaves as a proton acceptor with respect to water solution. Consequently, in solutions of acetic acid alone, the molar concentration of the HC,H,O is much larger than the C,H,O, concentration, but in solutions of sodium acetate alone, the reverse is truc. Neither solute alone can provide "comparable" concentrations in solutions of both the weak acid and its conjugate base; thus neither an acetic acid solu- tion nor a sodium acetate solution is a buffer solution. A buffer solution, by definition must contain moderate concentrations of both species The following is true of buffers: The pH of a solution of a weak acid (if not extremely weak) is governed by the concentration of the acid and K, . · The pH l of a solution of a weak base is determined by the concentration of the weak base and K In a solution containing both a weak acid and a strong acid, both acids play a role in determining the pH of the solution. E in the solution containing both a weak acid and a strong acid, the concentration of he strong acid is relatively large, it will inhlbis the dissociation of the weak acid . .If in the soluti of the strong acid is re 223

Explanation / Answer

Part C.

Beaker# 1 : 20 ml of 0.6 M HOAc

moles of HOAc = 0.6 x 20 = 12 mmol

ml of 1 M NaOH = 2 ml

moles of NaOH = 2 mmol

[HOAc] remaining = 10 mmol/22 ml = 0.454 M

[NaOAc] formed = 2 mmol/22 ml = 0.091 M

This is a buffer solution,

pH = pKa + log(base/acid)

      = 4.74 + log(0.091/0.454)

      = 3.31

Beaker#  2 : 20 ml of 0.6 M HOAc

moles of HOAc = 0.6 x 20 = 12 mmol

ml of 1 M NaOH = 5 ml

moles of NaOH = 5 mmol

[HOAc] remaining = 7 mmol/25 ml = 0.28 M

[NaOAc] formed = 5 mmol/25 ml = 0.2 M

pH = pKa + log(base/acid)

      = 4.74 + log(0.2/0.28)

      = 0.693

Beaker#  3 : 20 ml of 0.6 M HOAc

moles of HOAc = 0.6 x 20 = 12 mmol

ml of 1 M NaOH = 10 ml

moles of NaOH = 10 mmol

[HOAc] remaining = 2 mmol/30 ml = 0.066 M

[NaOAc] formed = 10 mmol/30 ml = 0.33 M

pH = pKa + log(base/acid)

      = 4.74 + log(0.33/0.066)

      = 5.44

Beaker#  4 : 20 ml of 0.6 M HOAc

moles of HOAc = 0.6 x 20 = 12 mmol

ml of 1 M NaOH = 20 ml

moles of NaOH = 20 mmol

Equivalence point

[NaOAc] formed = 20 mmol/40 ml = 0.5 M

Kb = 1 x 10^-14/1.8 x 10^-5 = x^2/0.5

x = [OH-] = 1.66 x 10^-5 M

pOH = -log[OH-] = 4.78

pH = 14 - pOH = 9.22

Part D - Addition of 3 drops of HCl

Beaker# 1

initial [HOAc] = 0.6 x 10/20 = 0.3 M

Ka = 1.8 x 10^-5 = x^2/0.3

x = [H+] = 2.32 x 10^-3 M

Initial pH = -log[H+] = 2.63

After HCl = 3 drops = 0.15 ml of 6 M = 6 x 0.15 = 0.9 mmol added

final [H+] = 0.9 mmol/20.15 ml = 0.044 M

pH = -log[H+] = 1.35

change in pH dpH = 2.63 - 1.35 = 1.28

Beaker#2 : initial [NaOAc] = 0.6 M x 10 ml/20 ml = 0.3 M

Kb = 1 x 10^-14/1.8 x 10^-5 = x^2/0.3

x = [OH-] = 1.29 x 10^-5 M

pOH = -log[OH-] = 4.89

pH = 14 - pOH = 9.11

After HCl = 3 drops = 0.15 ml of 6 M = 6 x 0.15 = 0.9 mmol added

[NaOAc] remains = (6 - 0.9)/20.15 ml = 0.25 M

[HOAc] formed = 0.9/20.15 = 0.044 M

final pH = 4.74 + log(0.25/0.044) = 5.49

change in pH dpH = 9.11 - 5.49 = 3.62

Part D - Addition of 3 drops of NaOH

Beaker# 1

initial [HOAc] = 0.6 x 10/20 = 0.3 M

Ka = 1.8 x 10^-5 = x^2/0.3

x = [H+] = 2.32 x 10^-3 M

Initial pH = -log[H+] = 2.63

After NaOH = 3 drops = 0.15 ml of 6 M = 6 x 0.15 = 0.9 mmol added

final [HOAc] remain = 5.1 mmol/20.15 ml = 0.25 M

Formed [NaOAc] = 0.9/20.15 = 0.044 M

final pH = 4.74 + log(0.044/0.25) = 3.98

change in pH = (initial - final)pH = -1.38

Beaker#2 : initial [NaOAc] = 0.6 M x 10 ml/20 ml = 0.3 M

Kb = 1 x 10^-14/1.8 x 10^-5 = x^2/0.3

x = [OH-] = 1.29 x 10^-5 M

pOH = -log[OH-] = 4.89

pH = 14 - pOH = 9.11

After NaOH = 3 drops = 0.15 ml of 6 M = 6 x 0.15 = 0.9 mmol added

final [OH-] = 0.9/20.15 = 0.044 M

pOH = -log[OH-] = 1.35

pH = 12.65

Change in pH dpH = 9.11 - 12.65 = -3.54

Beaker# 3 : Salt of strong acid and strong base

pH = 7

Beaker#4 : [HOAc] remain = ((0.6 x 10) - 3 mmol)/20 ml = 0.15 M

[NaOAc] formed = 3 mmol/20 ml = 0.15 M

pH = pKa = 4.74

Beaker#5 : 20 ml H2O

pH = 7

I have done calculations for beaker#1 and beaker#2 for part D. Similarly others can be done and calculate change in pH.

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