The salt formed by the reaction of the weak acid hydrocyanic acid, hcn, with the strong base potassium hydroxide is
potassium cyanide, kcn. what is the hydroxide ion concentration of a 0.255 m solution of potassium cyanide at 25 °c
given that the value of for hydrocyanic acid is 4.9 x 10-102

2.28 × 10^-3 mol/L

Explanation:

The equation for the equilibrium is

CN^- + H2O ⇌ HCN + OH^-

                    Ka = 4.9 × 10^-10

               KaKb = Kw

4.9 × 10^-10 Kb = 1.00 × 10^-14

                   Kb = (1.00 × 10^-14)/(4.9 × 10^-10) = 2.05 × 10^-5

Now, we can set up an ICE table

                     CN^- + H2O ⇌ HCN + OH^-

I/(mol/L)      0.255                     0         0

C/(mol/L)       -x                        +x        +x

E/(mol/L)  0.255 - x                   x         x

Ka = x^2/(0.255 - x) = 2.05 × 10^-5

Check for negligibility

0.255/(2.05 × 10^-5) = 12 000 > 400. ∴ x ≪ 0.255

    x^2 = 0.255(2.05 × 10^-5) = 5.20 × 10^-6

        x = sqrt(5.20 × 10^-6)    = 2.28 × 10^-3

[OH^-] = x mol/L                     = 2.28 × 10^-3 mol/L

it takes on an electronical energy

the answer is 0.8m.

by: nicholas