To understand how buffers use reserves of conjugate acid and conjugate base to a

Question

To understand how buffers use reserves of conjugate acid and conjugate base to ameliorate the effects of acid or base addition on pH. A buffer is a mixture of a conjugate acid-base pair in other words, it is a solution that contains a weak acid and its conjugate base or a weak base and its conjugate acid. For example, and acetic acid buffer consists of equiangular amounts of acetic acid, CH_3COOH, and its conjugate bases, CH_3COO^-. Because ions cannot empty be added to a solution, the conjugate base is added in a salt form (e.g., NaCH_3COO). Buffers work because the conjugate acid-base pair work together to neutralize the addition of H^+ or OH^- ions. These, for example, if H^+ ions are added to the acetate buffer described above, they will be largely removed from solution by the reaction of H^+ with the conjugate base. H^+ + CH_3COO^- CH_3COOH Similarly any added OH^- ions will be neutralized by a reaction with the conjugate acid. OH^+ + CH_3COOH + CH_3COO^- + H_2O This buffer system is described the equation PH = pK_a + log [conjugate based]/[conjugate acid] Part A A beaker with 145 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchlop. The total molarity of acid and conjugate based in this buffer is 0.100 M. A student adds 8.70 mL of a 0.280 M HCI solution to the beaker. How much will the pH change? The pK_a of acetic acid is 4.700.

Explanation / Answer

SOLUTION:

pH = pKa + log[CH3COO-] / [CH3COOH]

5 = 4.7 + log[CH3COO-] / [CH3COOH]

or log[CH3COO-] / [CH3COOH] = 5 - 4.7 = 0.3

Or [CH3COO-] / [CH3COOH] = 1.99

[CH3COO-] = 1.99 [CH3COOH] ========1)

Further [CH3COO-] + [CH3COOH] = 0.1M

or 1.99 [CH3COOH] +  [CH3COOH] = 0.1

2.99 [CH3COOH] = 0.1

[CH3COOH] = 0.1 / 2.99 = 0.033 M

Hence

Or [CH3COO-] = 0.1 - [CH3COOH]

[CH3COO-] = 0.1 - 0.033 = 0.067M

The HCl added will react with base CH3COO-

Moles of HCl added - molarity X volume = 0.2 X 8.7 = 1.74 mmol

Moles of CH3COO- = 0.067 X 145 = 9.715 mmoles

Similarly moles of CH3COOH = 145 X 0.033 = 4.785 mmoles

Out of 9.715 mmoles of CH3COO- , 1.74 mmol will react with HCl to form 1.74 mmol of CH3COOH as:

CH3COO- + HCl ----> CH3COOH + Cl-

Hence moles of CH3COO- that will remain after sddition of HCl = 9.715 - 1.74 = 7.975 mmol = 0.007976 Moles

Moles of CH3COOH (after addition of HCl) = 4.785 + 1.74 = 6.525 mmoles = 0.006525 Moles

Molarity of CH3COOH = Number of moles / Volume in liters

Volume = 145mL + 8.7mL = 153.7mL = 0.1537 L

Molarity of CH3COOH = 0.006525 Moles/ 0.1537M = 0.042 M

Similarly Molarity of CH3COO- = 0.00797 Moles / 0.1537 L = 0.052 M

Hence new pH (after addition of HCl) = 4.7 + log [0.052] /[0.042]

pH = 4.7 + 0.0899 = 4.789

Change in pH = 5 - 4.789 = 0.211

Hence pH change = - 0.211

Negative sign indicates decrease in pH

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