A buffer system is made of 0.250M HA (Ka = 5.75 x 10^-4) and 0.125M NaA.

Question

a) find pH

b) prove wether a buffer or not

c) what is pH after 0.100M HCL is added

Answer 1

You can use the Henderson Hasselbalch equation of the equilibrium constant equation. I prefer the HH.

a) pH = pKa + log [salt]/[acid]

First, find pKa, i.e. -log Ka. pKa = -log 5.75x10-4 = 3.24

pH = 3.24 + log [0.124]/[0.250] = 3.24 + (-0.30)

pH = 2.94

b) A buffer is a weak acid + the salt of that acid, or a weak base + the salt of that base. Here we have the weak acid HA and the salt of that acid, NaA, so it would be a buffer. Further proof is that if you add acid, it will react with the NaA to form the weak acid HA and since it's weak it won't ionize greatly, thus pH will not change much. If base is added, it reacts with HA to form the salt, and thus pH won't change much.

c) Adding 0.1 M HCl and assuming the volume doesn't change, it will react with the NaA as follows:

A- + H+ ==> HA so [HA] increases by 0.1 to give 0.350 M and [A-] decreases by 0.1 to give 0.025 M. Then,

pH = 3.24 + log [0.025/0.350] = 3.24 + (-1.15)

pH = 2.09