Learning Goal: To understand how buffers use reserves of conjugate acid and conj

Question

Learning Goal:

To understand how buffers use reserves of conjugate acid and conjugate base to counteract 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, an acetic acid buffer consists of acetic acid, CH3COOH, and its conjugate base, the acetate ion CH3COO-. Because ions cannot simply be added to a solution, the conjugate base is added in a salt form (e.g., sodium acetate NaCH3COO).

Buffers work because the conjugate acid-base pair work together to neutralize the addition of H+ or OH- ions. Thus, 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++CH3COO--->CH3COOH

Similarly, any added OH? ions will be neutralized by a reaction with the conjugate acid:

OH-+CH3COOH --> CH3COO-+H2O

This buffer system is described by the Henderson-Hasselbalch equation

pH=pKa+log[conjugate base]/[conjugate acid]   

A beaker with 125mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 5.00mL of a 0.380M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.

Express your answer numerically to two decimal places. Use a minus (-) sign if the pH has decreased.

Explanation / Answer

Answer

For the initial buffer solution:

[CH3COOH] + [CH3COO-] = 0.100

[CH3COO-] = 0.100 - [CH3COOH]

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

5.00 = 4.760 + log ([CH3COO-]/[CH3COOH])

0.24 = log ([CH3COO-]/[CH3COOH])

[CH3COO-]/[CH3COOH] = 100.24 = 1.738

(0.100 - [CH3COOH])/[CH3COOH] = 1.738

[CH3COOH] = 0.03652 M

[CH3COO-] = 0.100 - 0.03652 = 0.06348 M

After adding HCl:

Moles of HCl added = moles of CH3COOH formed = moles of CH3COO- lost

= volume x [HCl]

= 7.90/1000 x 0.370 = 0.002923 mol

Initial moles of CH3COOH = volume x [CH3COOH]

= 100/1000 x 0.03652 = 0.003652 mol

Initial moles of CH3COO- = volume x [CH3COOH]

= 100/1000 x 0.06348 = 0.006348 mol

Final moles of CH3COOH = 0.003652 + 0.002923 = 0.006575 mol

Final moles of CH3COO- = 0.006348 - 0.002923 = 0.003425 mol

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

= pKa + log (final moles of CH3COO-/final moles of CH3COOH) since volume is the same

= 4.76 + log (0.003425/0.006575) = 4.48

pH change ?pH = final pH - initial pH

= 4.48 - 5.00 = -0.52

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