To learn how to calculate ion concentrations in an aqueous solution of a strong diprotic acid.

sulfuric acid, h2so4, is a strong acid. its complete dissociation in aqueous solution is represented as

h2so4→h++hso4−

a hso4− anion can dissociate further by

hso4−⇌h++so42−

but the extent of dissociation is considerably less than 100%. the equilibrium constant for the second dissociation step is expressed as

ka2=[h+][so42−][hso4−]=0.012

part a

calculate the concentration of h+ ions in a 0.010 mol l−1 aqueous solution of sulfuric acid.

express your answer to three decimal places and include the appropriate units.

[h+] =

The concentration of H⁺ ions is 0.0165 M.

Explanation:

Let's consider the dissociation of H₂SO₄. In the first step, H₂SO₄ acts as a strong acid, completely dissociating into HSO₄⁻ and H⁺. Therefore, the concentrations of these ions will be the same that the initial concentration of the acid.

           H₂SO₄ ⇒ HSO₄⁻ +    H⁺

Initial   0.010M        0            0

Final         0        0.010M   0.010M

Now, HSO₄⁻ is a weak acid that will dissociate partially to form H⁺ and SO₄²⁻.

HSO₄⁻ ⇄ H⁺ + SO₄²⁻

To find out the concentration of H⁺ from HSO₄⁻ we will use an ICE Chart. We recognize 3 stages: Initial, Change and Equilibrium, and complete each row with the concentration or change in concentration.

    HSO₄⁻ ⇄ H⁺ + SO₄²⁻

I   0.010        0        0

C     -x          +x      +x

E  0.010 -x    x         x

This quadratic equation has 2 solutions: x₁ = -0.018 and x₂ = 0.00649. Since concentrations cannot be negative, we choose x₂. Then, [H⁺] coming from HSO₄⁻ is 0.00649 M.

The total concentration of H⁺ is:

[H⁺] = 0.010 M + 0.00649 M = 0.0165 M

organisms are more likely to be more preserved in mud than sand because mud is moist and wet, while sand is very dry.

the answer is 0.8m.

by: nicholas