Question

pH of acetic acid/acetate buffer pH of pure water Pure solution (no additions) Addition of 10 mL of 0.01 M HCI Addition of 10 mL of 0.01 M NaOH Comment on the changes in pH in the acetate buffer as opposed to that of pure water. What does this tell you about the role of a buffer?

Answer 1

A.)

Consider what would happen to the pH of a 0.5 M acetic acid (a weak acid) solution but which also contains 0.5 M sodium acetate (a salt of a weak acid).

The appropriate equation to describe this solution is:

CH₃COOH + H₂O <==> H₃O⁺ + CH₃COO^-

The pH of the solution can be calculate if you know the K_a of acetic acid = 1.8 x 10^-5. Plug this into the following equation to get x = [H₃O⁺] and pH as follows:

K_a = ([H₃O⁺] [CH₃COO^-])/[CH₃COOH] = x(0.5)/0.5 = 1.8 x 10^-5

x = [H₃O⁺] = 1.8 x 10^-5; pH = 4.74

Now if 0.01 mol of HCl was added (without changing the volume), the HCl would react with the weak base acetate as shown below:

HCl + CH₃COO^- <==> Cl^- + CH₃COOH

Therefore the added HCl is consumed, CH₃COO^- is reduced to 0.49 M and CH₃COOH increases to 0.51. As above, x can be calculated and the pH determined.

K_a = ([H₃O⁺] [CH₃COO^-])/[CH₃COOH] = x(0.49)/0.51 = 1.8 x 10^-5

x = [H₃O⁺] = 1.87 x 10^-5; pH = 4.73. The pH was reduced only by 0.01 units in comparison to the case of pure water whose pH dropped by 5 units.

Now if 0.01 mol of NaOH was added (without changing the volume), the hydroxide would react with the weak acid as shown below:

CH₃COOH + OH^- <==> H₂O + CH₃COO^-

Therefore the base is consumed, CH₃COOH is reduced to 0.49 M and CH₃COO^- increases to 0.51. As above, x can be calculated and the pH determined.

K_a = ([H₃O⁺] [CH₃COO^-])/[CH₃COOH] = x(0.51)/0.49 = 1.8 x 10^-5

x = [H₃O⁺] = 1.72 x 10^-5; pH = 4.76.

The pH was raised only by 0.02 units in comparison to the case of pure water whose pH increased by 5 units.

Here are some diagrams which will give you a better idea.

A. 0.50 M acetic acid and 0.5 M acetate K,-([H30+] [acet atel)/acetic acid 1.8 x10-3: x(O5)/05 x [H 01 18 x 103 pH = 4 .74 pH

C. Now add 0.01 mol HCI to 1 L of A 11-CH,CooCH.CO0 K. = (H30+] [acet ate])/acetic acid 1.8 x10-5: x(049)/0.51 x [H301 187 x 10-5 pH 4.73 pH

B. Now add 001 mol N aOH to 1 Lof A 11CH,COOHOH > H20CH2CO0 K. = ([H30+] [ acetatell/acetic acid 91.8 x 103 x(051)/0.49 x = [H301-172 x 105 pH-4.76 PH 7

B.)

Consider what would happen to the pH of pure water (pH = 7) if either 0.01 mol of NaOH or 0.01 mol of HCl were added to sufficient water to make 1L. The first solution would now have an [OH^-] = 0.01 M and hence a [H₃O⁺]=1x10^-12 M, or a pH = 12. The second solution would have a [H₃O⁺] = 1 x 10^-2 M, or a pH = 2. This examples shows that the addition of a small amount fo either acid or base to water dramatically changes the pH. Consider the pH of pure water to be like a balance. Addition of small amounts of either acid or bases dramatically changes the balance (pH). See the pictures below, it will give you an idea:

B. Add to 0.01 mole NaOH to 1 L of above H20OH20OH [OH-] = 0.01 M: [H30+] = 1 X 10-12 M pH = 12 9 pH

C. Add 0.01 mole HCI to A. H30++ Ch [H301-001 M pH 2 9 pH 5