Question

You have 325 mL of an 0.55 M acetic acid solution. What volume (V) of 1.80 M NaOH solution must you add in order to prepare an acetate buffer of pH = 5.38? (The pKa of acetic acid is 4.76.)

Answer 1

pH of the buffer is given by Henderson- Hasselbatch equation

pH= pKa+log [A-]/[HA]

5.38= 4.76 + log[A-]/[HA]

[A-]/[HA] = 10^(5.38 - 4.76)= 4.16

initial [HA] = 0.55 M

moles of HA=4.16*0.55= 2.288

After adding a volume V of 1.80 M NaOH, moles of HA reduce by 1.80*V and increase moles of A- by the same amount.

volume of new solution is 4.16 + V so the new concentrations are

[HA] = (2.288 - 1.80*V)/(0.325 + V)

[A-] = 1.80*V/(0.325 + V)

[A-]/[HA] = 1.80*V/(0.325 - V)

4.16= 1.80*V/(0.325 - V)

V= 0.226 L= 226 ml