Answer:
pH = 0.18
Explanation:
The sulfuric acid (H₂SO₄) has the following reactions in aqueous medium:
H₂SO₄ → HSO₄⁻ + H⁺
HSO₄⁻ ⇄ SO₄²⁻ + H⁺ Ka = 1.1x10⁻²
Where Ka is defined as Ka = [SO₄²⁻] [H⁺] / [HSO₄⁻] = 1.1x10⁻²
Based in the first reaction, [H⁺] = 0.65M and [HSO₄⁻] = 0.65M
In the second reaction, the two species are in equilibrium, thus, concentrations will be:
[H⁺] = 0.65M + X
[HSO₄⁻] = 0.65M - X
[SO₄²⁻] = X
Replacing in Ka formula:
1.1x10⁻² = [X] [0.65 + X] / [0.65M - X]
7.15x10⁻³ - 1.1x10⁻²X = 0.65X + X²
0 = X² + 0.661X - 7.15x10⁻³
Solving for X:
X = -0.67M → False solution. There is no negative concentrations.
X = 0.01065M → Right answer.
Thus [H⁺] = 0.65M + 0.01065M = 0.66065M
As pH = -log [H⁺];
pH = -log 0.66065M = 0.18
