Find the hydrogen ion, H+, and hydroxide ion, concentration in 0.02 mole solution of hydrogen tetraoxosulphate(vi)

1 answer

To find the concentration of the hydrogen ion (H+) and hydroxide ion (OH-) in a 0.02 mole solution of hydrogen tetraoxosulphate(vi) (H2SO4), we need to first identify the dissociation of the acid.

H2SO4 + H2O → 2H+ + SO4^2-

Since H2SO4 is a strong acid, it will dissociate completely in water.

From the balanced equation, the mole ratio between H2SO4 and H+ is 1:2. This means that for every mole of H2SO4 that dissociates, 2 moles of H+ are produced.

Given that the concentration of H2SO4 is 0.02 moles, the concentration of H+ ions will be twice that amount:

Concentration of H+ = 2 * 0.02 moles = 0.04 moles

Since the solution is in 1 liter of water, the concentration of H+ ions is:

Concentration of H+ ions = 0.04 moles / 1 L = 0.04 moles/L

The concentration of the hydroxide ion (OH-) can be determined using the equation Kw = [H+][OH-], where Kw is the ionization constant of water (1 x 10^-14 at 25°C).

Assuming the solution is at 25°C, then the concentration of OH- ions can be calculated as follows:

Kw = [H+][OH-]
1 x 10^-14 = 0.04 moles/L * [OH-]
[OH-] = 1 x 10^-14 / 0.04 ≈ 2.5 x 10^-13 moles/L

Therefore, the concentration of H+ ions is 0.04 moles/L, and the concentration of OH- ions is approximately 2.5 x 10^-13 moles/L in a 0.02 mole solution of hydrogen tetraoxosulphate(vi).