Find the equilibrium concentration of H2SO3(aq)

Question

The pH of a 0.401 mol/L solution of sulfurous acid is 1.18. Based on this evidence, the equilibrium concentration of H₂SO_3(aq) is a.bc mol/L.

Answer 1

We need to determine concentration based on the equilibrium reaction:

H₂SO₃ ↔ H⁺ + HSO₃^-

Basically, the acid just loses a proton. It becomes harder and harder to rip of hydrogen atoms as the sulfurous base becomes more electronegative, so this is where we're stuck at. We are told that the pH of the solution is 1.18. We can determine the concentration of [H⁺] using this pH and the equation:

pH = -log[H⁺]

to get

1.18 = -log[H⁺]

Solve for H⁺ concentration by isolating H⁺ (assuming you know log rules). You should get:

[H⁺] = 10^-1.18

This gives an H⁺ concentration of 0.066 M H⁺. What does this mean? This means that the sulfurous acid that started at 0.401 M has given up 0.066 M to create the products. (It's like if you have $40 but only spent $6. To find out how much is left, we subtract.) If it's the deprotonated product, its no longer sulfurous acid -so we must subtract it from the original concentration:

NEW H₂SO₃ concentration: 0.401 - 0.066 = 0.335 M H₂SO₃

a = 0

b = 3

c = 4