Acid/Base equilibrium calculations

Question

A solution contains 0.308 M HA (K_a = ) and 0.695 M NaA.  

What is the pH of this solution after 0.192 mol of HCl are added to 1.00 L of this solution?

Answer 1

This is a solution of a weak acid (HA) and the salt or conjugate base of that acid (NaA). This makes a BUFFER. We can use the Henderson Hasselbalch equation to find pH:

pH = pKa + log [salt]/[acid]. The pKa = -log Ka = -log 7.45x10^-7 = 6.55

Next we must find the [salt] and [acid] after the addition of 0.192 mol HCl.

NaA + HCl ===> HA + NaCl

0.695....0.192.........0.308.............Initial

-0.192...-0.192.......+0.192...........Change

0.503.......0............0.500............Equilibrium

Now we can plug these values into the HH equation and solve for pH...

pH = 6.55 + log (0.503/0.500)

pH = 6.55 + log 1.006

pH = 6.55 + 0.0026

pH = 6.5526 = 6.55