Question

Calculate the theoretical buffer capability of 0.1 M CH3COOH/CH3COONa buffer at pH=4.74 and pH=6 (I don't know does the pH value affect so I provide it anyway)

pKa of acetic acid is 4.74

To calculate the buffer capability, meansure the change by adding 0.5mL of 1.0M HCl and by adding 0.5mL NaOH to 10 mL of buffer, the capability is then the determined as the molarity of the HCl or NaOH added to the buffer divided by the change of pH.

Answer 1

Buffer capacity is a measure of efficiency of buffer to resist changes in pH. Buffer capacity () is expressed asamount of acid or base in gram equivalents that must be added to 1 litre of buffer solution to change its pH by one unit.

= B / pH

B = gram equivalents of acid or base that must be added to 1 litre of buffer solution

pH = change in pH caused by addition of acid or base.

pH = pK_a + log [A^-]/ [HA]

pH = 4.74 + log 0.1

pH = 4.74 - 1

pH = 3.74

When we add 1M HCl in a litre of solution then there is change in concentration of buffer solution the new concentration would be

10 ml of buffer means 0.001 l of buffer so 1 X 0.001 = 0.001moles of HCl

0.1M of buffer - 0.001M of HCl = 0.099M and 0.1 + 0.001 = 0.101M

New pH

pH = 4.74 + log 0.099 / 0.101

pH = 4.74 + log0.9801

pH = 4.74 - 0.0087

pH = 4.73

So there would be very small change in pH when HCl added to the buffer solution pH = 3.74 - 4.73 = -0.99

Gram equivalent weight of 1 mol of HCl contains= 1 + 35.45 = 36.45

0.099 mol contains = 36.45 X 0.099 = 3.61

0.101 mol contains = 36.45 X 0.101 = 3.68

B = 3.68 - 3.61 = 0.07

= 0.07/ -0.99 = - 0.92

Now by adding NaOH

pH = pK_a + log [A^-]/ [HA]

pH = 4.74 + log 0.1

pH = 4.74 - 1

pH = 3.74

When we add 1M NaOH in a litre of solution then there is change in concentration of buffer solution the new concentration would be

10 ml of buffer means 0.001 l of buffer so 1 X 0.001 = 0.001moles of NaOH

0.1M of buffer - 0.001M of NaOH = 0.099M and 0.1 + 0.001 = 0.101M

New pH

pH = 4.74 + log 0.099 / 0.101

pH = 4.74 + log0.9801

pH = 4.74 - 0.0087

pH = 4.73

So there would be very small change in pH when NaOH added to the buffer solution pH = 3.74 - 4.73 = -0.99

Gram equivalent weight of 1 mol of NaOH contains= 23 + 1 + 16 = 40

0.099 mol contains = 40 X 0.099 = 3.96

0.101 mol contains = 40 X 0.101 = 4.04

B = 4.04 - 3.96 = 0.08

= 0.08/ -0.99 = - 0.081