You have 175 mL of an 0.51 M acetic acid solution. What volume (V) of 1.10 M NaOH solution must you add in order to prepare an acetate buffer of pH = 5.01? (The pKa of acetic acid is 4.76.)
Here the volume = 175 ml = 0.175 L
First calcaulte the ratio of salt and acid as follows:
pH = pKa + log[A-]/[HA]
5.01 = 4.76 + log[A-]/[HA]
[A-]/[HA] = 10^(0.25) = 1.78
initial [HA] = 0.51M
initial moles of HA = 0.175*0.51 = 0.08925 mol
after adding a volume V of 1.10 M NaOH, we will reduce moles of HA
by 1.10*V and increase moles of A- by the same amount.
The new solution volume is 0.175 + V so the new concentrations
are
[HA] = (0.08925 - 1.10*V)/(0.175+ V)
[A-] = 1.10*V/(0.175+ V)
[A-]/[HA] = 1.10*V/(0.175 - V)
1.78 = 1.10*V/(0.175- V)
V = 0.244 L = 244 mL
Get Answers For Free
Most questions answered within 1 hours.