Question

A chemist titrates 120.0 mL of a 0.4006 M hydrocyanic acid (HCN) solution with 0.6812 MNaOH solution at 25 °C. Calculate the pH at equivalence. The pKa of hydrocyanic acid is 9.21.

Answer 1

Answer:

pH = 11.30

Explanation:

The reaction of HCN with NaOH is:

HCN + NaOH → H₂O + Na⁺ + CN⁻

Where 1 mole of hydrocyanic acid reacts with 1 mole of NaOH.

Moles of HCN are:

0.1200L × (0.4006mol / L) = 0.04807moles HCN

Thus, at equivalence, moles of NaOH you must add are 0.04807moles. And you must add:

0.04807moles × (1L / 0.6812mol) = 0.07057L of the 0.6812M NaOH solution.

Concentration of CN⁻ at equivalence point is:

0.04807moles / (0.1200L + 0.07057L) = 0.252M CN⁻

This, CN⁻ is in equilibrium with water thus:

CN⁻(aq) + H₂O(l) ⇄ HCN(aq) + OH⁻(aq)

Where Kb is:

Kb = [HCN][OH⁻] / [CN⁻] (1)

Kb = Kw / Ka

Ka is -log pKa → 6.166x10⁻¹⁰

Kb = 1x10⁻¹⁴ / 6.166x10⁻¹⁰ = 1.62x10⁻⁵

Concentrations in equilibrium are:

[CN⁻] = 0.252M - X

[HCN] = X

[OH⁻] = X

Replacing in (1):

1.62x10⁻⁵ = [X][X] / [0.252M - X]

4.08x10⁻⁶ - 1.62x10⁻⁵X = X²

0 = X² +  1.62x10⁻⁵X - 4.08x10⁻⁶

Solving for x:

X = -0.002 → False answer. There is no negative concentrations.

X = 0.00201 → Right answer.

As [OH⁻] = X; [OH⁻] = 0.00201

pOH is -log  [OH⁻]

pOH = 2.70

As 14 = pOH + pH

pH = 11.30