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

I’d have to say no because without oxygen in the water the fish would all be dead. so if you want to fish for dead fish, pick the ponds with the lowest oxygen concentrations.

A.