Question

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

H2SO3 ---> H+ + HSO3-
HSO3- ---> H+ + SO32-
If .50 moles of sulfurous acid are dissolved to form a 1 L solution, which of the following concentrations will be LEAST at equilibrium?
A.) [H2SO3]
B.) [H+]
C.) [H3O+]
D.) [SO32-]

Answer 1

The [SO₃²⁻]

Step-by-step explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

`K_(a1) = ([H^(+)] [HSO_(3)^(-)])/([H_(2)SO_(3)]) = (x\cdot x)/(0.5 - x) = \frac {x^(2)}{0.5 -x}`

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

`[HSO_(3)^(-)] = [H^(+)] = 7.94 \cdot 10^(-2)M`

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

`K_(a2) = ([H^(+)] [SO_(3)^(2-)])/([HSO_(3)^(-)]) = (x^(2))/(7.94 \cdot 10^(-2) - x)`

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

`[SO_(3)^(2-)] = [H^(+)] = 7.07 \cdot 10^(-5)M`

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!