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HOCN(aq)⇄H+(aq)+OCN−(aq)Ka=3×10−4

The ionization of cyanic acid, HOCN, is represented above. A certain solution of HOCN has a pH of 2.0. What is the approximate concentration of unionized HOCN(aq) in the solution?


3×10^−4M

1×10^−2M

3×10^−1M

2×10^0M



Answer :

pstnonsonjoku

The concertation of the undissociated acid is 3×10^−1M.

What is the concentration of the unionized acid?

Now let us know that the acid dissociation constant shows the extent to which the acid would be dissociated in a given time. To obtain the concentration of the unionized acid, we need to first obtain the the concentration of the hydrogen ion.

Using;

pH = 2

[H^+] = Antilog (-2) = 0.01 M

Ka = [H^+] [OCN−]/HOCN

Given that  [H^+] = [OCN−] = 0.01 M

Ka = [H^+]^2/HOCN

HOCN =  [H^+]^2/3×10^−4

HOCN =   (0.01)^2/3×10^−4

HOCN =  3×10^−1M

Learn more about acid dissociation:https://brainly.com/question/15012972

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