Answer:
pH = 11.30
Explanation:
The reaction of HCN with NaOH is:
HCN + NaOH → H₂O + Na⁺ + CN⁻
<em>Where 1 mole of hydrocyanic acid reacts with 1 mole of NaOH.</em>
<em> </em>
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) = <em>0.252M CN⁻ </em>
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This, CN⁻ is in equilibrium with water thus:
CN⁻(aq) + H₂O(l) ⇄ HCN(aq) + OH⁻(aq)
Where Kb is:
Kb = [HCN][OH⁻] / [CN⁻] <em>(1)</em>
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
<em>pH = 11.30</em>
