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