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.

1. they vary with a higher number because boiling is way hotter than melting