What is the ph of a buffer prepared by adding 0.607 mol of the weak acid ha to 0.609 mol of naa in 2.00 l of solution? the dissociation constant ka of ha is 5.66*10-7?

Given:

0.607 mol of the weak acid

0.609 naa

2.00 liters of solution

The solution for finding the ph of a buffer:

[HA] = 0.607 / 2.00 = 0.3035 M

[A-] = 0.609 / 2.00 = 0.3045 M

pKa = 6.25

pH = 6.25 + log 0.3045 / 0.3035 = 6.25 is the ph buffer prepared.