Calculate the concentration and pH of a M aqueous solution of sodium cyanide, . Finally, calculate the concentration. ()

[OH⁻] = 0.000324619M

pH = 10.51

[CN⁻] = 5.175x10⁻³M

Explanation:

Full question says:

Calculate the OH- concentration and pH of a 5.5×10-3M aqueous solution of sodium cyanide, NaCN. Finally, calculate the CN- concentration.

Ka (HCN) = 4.9×10-10

Sodium cyanide, is in equilibrium with water as follows:

NaCN(aq) + H₂O(l) ⇄ HCN(aq) + OH⁻(aq)

Where Kb (Kb = Kw / Ka = 1x10⁻¹⁴ / 4.9x10⁻¹⁰ = 2.04x10⁻⁵) is:

2.04x10⁻⁵ = [HCN] [OH⁻] / [NaCN]

In equilibrium, ann amount of HCN and OH⁻ is produced = X and the molarity of NaCN is: 5.5x10⁻³M - X

Replacing in Kb expression:

2.04x10⁻⁵ = [X] [X] / [5.5x10⁻³M - X]

1.12x10⁻⁷- 2.04x10⁻⁵X = X²

1.12x10⁻⁷- 2.04x10⁻⁵X - X² = 0

Solving for X:

X = -0.0003 → False solution. There is no negative concentrations

X = 0.000324619

As [OH⁻] = X.

pOH = -log [OH⁻]

pOH = 3.489

pH = 14 -pOH

And as molarity of NaCN = [CN⁻] = 5.5x10⁻³M - X

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