What is the concentration of h+ in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5×10−6 and ka2=1.2×10−11?

The equation of the first dissociation
H₂X(aq) + H₂O(l) ⇄ HX⁻(aq) + H₃O⁺(aq)
The acid dissociation is constant Kₐ1 is 4.5 × 10⁻⁶
To construct ICE table and equilibrium concentration
H₂X(aq) + H₂O(l) ⇄ HX⁻(aq) + H₃O⁺(aq)
I (M):   0.150        0   0
C (M):   ₋x            ₊x   +x
E (M):    0.150₋x    x     x
Acidiozation constant
4.5 × 10⁻⁶ = (x) (x)/(0.150 - x)
x = 0.000820M
From equilibrium table
[H₃O⁺] = [4x⁻] = x = 0.000820M
[H₂x] = (0.0150 - 0.000820) = 0.0143M
Equation of second dissociation
HX⁻(aq) + H₂O(l) ⇆ X⁻(aq) ₊ H₃O⁺(aq)
acid dissociation Ka₂ is 1.2 × 10⁻¹¹
The ICE table and equilibrium concentration
HX⁻(aq) + H₂O(l) ⇆ X⁻(aq) ₊ H₃O⁺(aq)
I(M):     0.000820    0     0.00820
C(M):     -Y                 +Y   +Y
E(M):    0.000820-Y     Y    0.000820+Y
Acidiozation constant
1.2 × 10⁻¹¹ = (y) (0.000820₊y/0.000820-y
y = 1.20 × 10⁻¹¹ M
From equilibrium table
[H₃O⁺] = 0.000820 + 1.20 × 10⁻¹¹
= 8.20 × 10⁻4 M

The answer is b i think