Question

In: Chemistry

Consider a 300mL sample of 0.250M HCN (Ka=4.9x10^-10). To find the pH, do you need an...

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)

Solutions

Expert Solution

[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  


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