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The acid-dissociation constant of hydrocyanic acid (hcn) at 25.0 °c is 4.9 ⋅ 10−10. what is the ph of an aqueous solution of 0.080 m sodium cyanide (nacn)?

Respuesta :

superman1987
According to the reaction equation:

and by using ICE table:

              CN-  + H2O ↔ HCN  + OH- 

initial  0.08                        0          0

change -X                        +X          +X

Equ    (0.08-X)                    X            X

so from the equilibrium equation, we can get Ka expression

when Ka = [HCN] [OH-]/[CN-]

when Ka = Kw/Kb

               = (1 x 10^-14) / (4.9 x 10^-10)

               = 2 x 10^-5

So, by substitution:

2 x 10^-5 = X^2 / (0.08 - X)

X= 0.0013

∴ [OH] = X = 0.0013 

∴ POH = -㏒[OH]

            = -㏒0.0013

            = 2.886 

∴ PH = 14 - POH

         = 14 - 2.886 = 11.11
pstnonsonjoku

The pH of the solution is calculated to be 5.22.

  • First we must set up the table for the reaction:

Let the cyanide ion be represented by X

       HX(aq) ⇄   H^+(aq) + X^-(aq)

I      0.08             0                0

C   -x                    +x              +x

E   0.08 - x           x                x

  • We now obtain the Ka of the solution as follows;

Ka = [H^+] [ X^-]/[HX]

4.9 × 10^−10 = x^2/ 0.08 - x

4.9 × 10^−10(0.08 - x) = x^2

3.92 × 10^11 - 4.9 × 10^−10x = x^2

x^2 + 4.9 × 10^-10x - 3.92 × 10^-11 = 0

x = 0.000006 M

  • Now, the pH of the solution is obtained from;

pH = - log(0.000006 M)

pH = 5.22

Hence, the pH of the solution is calculated to be 5.22.

Learn more about pH: https://brainly.com/question/491373

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