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)?

The pH of the solution is calculated to be 5.22.

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

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

pH = - log(0.000006 M)

pH = 5.22

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

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120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890