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-]

The [SO₃²⁻]

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:

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

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:  

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

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!

The evidence that shows that a chemical reaction has taken place is the disappearance of the steel wool shows that a new substance has formed.

The electrochemical series is arrangement of elements in order of decreasing reactivity. It is important to note that the elements that are high up in the electrochemical series can displace the elements that are lower in the series from their aqueous solution.

Steel contains iron which lies above copper in the electrochemical series hence a chemical change occurs when steel dissolves in the copper solution.

Hence, the evidence that shows that a chemical reaction has taken place is the disappearance of the steel wool shows that a new substance has formed.

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