In: Chemistry
Consider a 300mL sample of 0.250M HCN (Ka=4.9x10^-10). To find the pH, do you need an ice table, can you use the Henderson-Hasselbalch equation, or something else? If an ice table, will it be a Ka or Kb equilibrium? Can you use the small x assumption? Is this a buffer? If something else, then how?
(Need Help Please!!! If possible could you work it out so i can know how to do it)
[HCN] = 0.250 M
ICE table
HCN + H2O <==> H3O+ + CN-
Initial 0.250 - -
change -x +x +x
Eq (0.250-x) x x
Ka = [H3O+][CN-]/[HCN]
4.9 x 10^-10 = x^2/0.250
Since Ka is very small, extent of dissociation x would be small and can be ignored in the denominator,
4.9 x 10^-10 = x^2/0.250
x = [H3O+] = 1.107 x 10^-5 M
pH = -log[H3O+] = 4.956