In: Chemistry
What is the pH of 40.0 mL of a solution that is 0.15 M in CN– and 0.27 M in HCN? For HCN, use Ka = 4.9 × 10–9.
This is a buffer
pH = pKa + log(A-/HA)
pKA = -log(KA) = -log(4.9*10^-9 ) = 8.3098
[A-] = 0.15
[HA] = 0.27
then
pH = pKa + log(A-/HA)
pH = 8.3098 + log(0.15/0.27) = 8.0545