Question

What is the pH of a buffer prepared by adding 0.506 mol of the weak acid HA to 0.305 mol of NaA in 2.00  L of solution? The dissociation constant Ka of HA is 5.66*10^-7.

pH = __________
What is the \rm pH after 0.150 mol of \rm HCl is added to the buffer from Part A? Assume no volumechange on the addition of the acid.
pH = _____________
What is the \rm pH after 0.195 mol of \rm NaOH is added to the buffer from Part A? Assume novolume change on the addition of the base.
pH = ______________

Answer 1

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

pKa = -log(5.66*10^-7)
=7-log(5.66) = 6.247

[salt] = 0.305/2 = 0.152
[acid] = .506/2=.253

1. pH = 6.247 + log(.152/.253)
=6.247-0.221 =6.026

2.
if strong acid added it will nutrilize salt(conjugate base) and make HA
so new [HA] = 0.506+0.150/2 = 0.328
[NaA] = 0.305-0.150 /2 = 0.0775

so pH =
=6.247 - log(0.0775/0.328)
= 6.247-0.626
=5.621

3. stong base added it will make NaA from HA
so new
[HA] = 0.506-0.195 /2 = 0.155
[NaA] = 0.305+0.195/2=0.250

so pH = 6.247 + log (0.250/0.155)
=6.246+0.207 = 6.454

ANSWER