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A 0.250 M solution of HCN has a pH = 4.87.


a) Determine the [H3O+] in the solution.

b) Calculate the percent ionization of HCN in this solution.

Respuesta :

Answer:

a. [H+]= 1.35x10^-5 M

b.5.4x10^-3%

Explanation:

a) pH= - log [H+]

[H+]= 10^-pH       [H+]= 10^-4.87    [H+]= 1.35x10^-5 M

b) HCN  -> H  + CN        at equilibrium  [H+]=[CN-]

0.250 M      -> 100%

1.35x10^-5 M ->x

x=(1.35x10^-5 M * 100)/0.250M

x=5.4x10^-3%

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