a 10.0 ml sample of 0.200 m hydrocyanic acid (hcn) is titrated with 0.0998 m naoh. what is the pH at equivalence point?

Consider the titration of HCN against NaOH.
HCN(aq) + OH⁻(aq) → CN⁻(aq) + H₂O(l)
Mole ratio HCN : OH⁻ : CN⁻ = 1 : 1 : 1

No. of milli-moles of HCN = (0.200 mmol/mL) × (10.0 mL) = 2.00 mmol
No. of milli-moles of NaOH = 2.00 mmol
Volume of NaOH = (2.00 mmol) / (0.0998 mmol/mL) = 20.2 mL
No. of milli-moles of CN⁻ formed = 2.00 mmol
Molarity of CN⁻ formed = (2.00 mmol) / [(10.0 + 20.2) mL] = 0.0662 M

pKa(HCN) = 9.31
Then, pKb(CN⁻) = 14.00 - 9.31 = 4.69
Kb(CN⁻) = 10⁻⁴·⁶⁹

Consider the dissociation (hydrolysis) of CN⁻ ions :
CN⁻(aq) + H₂O(l) ⇌ HCN(aq) + OH⁻(aq) ...... Kb = 10⁻⁴·⁶⁹

Initial concentrations :
[CN⁻]ₒ = 0.0662 M
[HCN]ₒ = [OH⁻]ₒ = 0 M

Equilibrium concentrations :
[CN⁻] = (0.0662 - y) M ≈ 0.0662 M, assuming 0.0662 ≫ y
[HCN] = [OH⁻] = y M

Kb = [HCN] [OH⁻] / [CN⁻]
y² / 0.0662 = 10⁻⁴·⁶⁹
y = √(0.0662 × 10⁻⁴·⁶⁹) = 0.00116

pOH = -log[OH⁻] = -log(0.00116) = 2.94
pH = pKw - pOH = 14.00 - 2.94 = 11.06