Calculate the pH of 0.65 M H2SO4 (sulfuric acid) solution? (Ka2: 1.1x10-2)

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

here's the link to your question!

explanation:

c is the answer because when something is left after something evaporates that shows the chemical change.