General Chemistry 2

Question

The pH of an aqueous solution of 0.248 M sodiumcyanide, NaCN, is (answer). (Assume that Ka (HCN) = 4.00 x 10^-10).

This solution is _____ (acidic, neutral, basic).

Answer 1

To find the pH of this solution, we must look at the hydrolysis of the CN^- anion, as follows:

CN^- +H₂O ==> HCN + OH^-

The Kb expression for this is

Kb = [HCN][OH^-] / [CN^-]

The Kb for CN^- can be calculated from the Ka for HCN (given as 4.00x10^-10)

Ka Kb = Kw

Kb = Kw / Ka = 1x10^-14 / 4.00x10^-10

Kb = 2.50x10^-5

Plugging values into this Kb expression, we have ...

Kb = [HCN][OH^-] / [CN^-]

2.50x10^-5 = (x)(x) / 0.248

x² = 6.20x10^-5

x = 2.49x10^-3 = [OH^-]

pOH = -log 2.49x10^-3 = 2.60

pH = 14 - 2.60

pH = 11.40

This solution is __basic___ (acidic, neutral, basic).

Answer 2

The CN^- is the conjugate base of a weak acid, so it will combine with H⁺ to form HCN. This results in a basic solution.

Calculation: Let x be the reaction needed to reach equilibrium concentration values

CN^- + H₂O = HCN + OH^- K_B = K_W/K_A = 10^-14/(4 x 10^-10) = 2.5 x 10^-5

Initial: .248 , 0, 10^-7 (Usually smaller than x, so we set it to 0)

Change - x, x x

Equil: .248-x x 10^-7+x

K_B = [HCN][OH^-]/[CN^-] = x(10^-7+x)/(.248-x)

Assume .248>>x>>10^-7

x = sqrt(.248(2.5 x 10^-5)) = .0025 which satisfies assumption (5% - our x is 1% Conversion)

The pH = 14-pOH and pOH = -log(x)

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