can someone answer this chem problem please?

Let y be the pH after addition of Ba(OH)₂.

Equation for the reaction between Ba(OH)₂ and HY:
Ba(OH)₂ + 2HY → BaY₂ + 2H₂O
Each moles of Ba(OH)₂ reacts with 2 moles of HY to give 2 moles of Y⁻.

After addition of Ba(OH)₂:
Moles of HY = (0.113 × 0.350 - 0.0014 × 2) mol = 0.03675 mol
Moles of Y⁻ = (0.203 × 0.350 + 0.0014 × 2) mol = 0.07385 mol
[Y⁻]/[HY] = (Moles of Y⁻)/(Moles of HY) = 0.07385/0.03675

Consider the dissociation of HY:
HY(aq) + H₂O(ℓ) ⇌ H₃O⁺(aq) + Y⁻(aq)

Henderson-Hasselbach equation: pH = pKa + log([Y⁻]/[HY])
Then, pKa = pH - log([Y⁻]/[HY])

Before addition of Ba(OH)₂:
pKa = 8.94 - log(0.203/0.113) …… [1]

After addition of Ba(OH)₂:
pKa = y - log(0.07385/0.03675) …… [2]

[2] = [1]:
y - log(0.07385/0.03675) = 8.94 - log(0.203/0.113)
y = 8.94 - log(0.203/0.113) + log(0.07385/0.03675)
y = 8.99
pH after Ba(OH)₂ is added = 8.99