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Let B be the weak base, and y M be the initial concentration of B.

Equation for the reaction :
B(aq) + H₂O(l) ⇌ HB(aq) + OH⁻(aq)

pOH = pKw - pH = 14.00 - 10.00 = 4.00
[OH⁻] at equilibrium = 10^(-pH) M = 10⁻⁴ M
Therefore, 10⁻⁴ M of the base is consumed to form 10⁻⁴ M of HB and 10⁻⁴ M of OH⁻ ions.

____________ B(aq) + H₂O(l) ⇌ HB(aq) + OH⁻(aq) ___ Kc = 7.20 × 10⁻⁷
At eqm : __ (y + 10⁻⁴) M _____ 10⁻⁴ M __ 10⁻⁴ M

As Kc is very small, B dissociates to a very small extent.
Hence, we can assume that y ≫ 10⁻⁴ M, and thus (y - 10⁻⁴) M ≈ y M

At equilibrium :
Kc = [HB] [OH⁻] / [B]
7.20 × 10⁻⁷ = (10⁻⁴)² / y
y = (10⁻⁴)² / (7.20 × 10⁻⁷)
y = 0.014 (to 2 sig. fig.)
(The value of y agree with the assumption that y ≫ 10⁻⁴ M)

The concentration of the base = 0.014 M

pOH = 0.5(pKb - lg(HB)) = 4

lg(HB) = -1.857

(HB) = 1.39*10^-2 M = 0.014 M
HB: weak base