Chemistry 112

Step 1:

H₂SO₄(aq) + H₂O(l) → H₃O⁺(aq) + HSO₄^–(aq)

0.0050 M H₂SO₄ generates 0.0050 M H₃O⁺ and 0.0050 M HSO₄^–

Step 2:

HSO₄^–(aq) + H₂O(l) → ← H₃O⁺(aq) + SO₄^2–(aq)

K_a = [H₃O⁺]_e[SO₄^2–]_e                             [HSO₄^–]_e = 1.1×10^–2

Initial0.0050 0.0050 0

Change–x +x +x

Equilibrium0.0050 – x 0.0050 + x x

Can we approximate? 0.0050/0.011 = 0.45 NO!

1.1×10^–2 = (0.0050 + x)(x)                        (0.0050 – x)

5.5×10^–5 – 1.1×10^–2x = 5.0×10^–3x + x²

x² + 1.6×10^–2x – 5.5×10^–5 = 0

We will need to use the quadratic equation to give:

x = 2.9×10^–3 or x = –1.9×10^–2

Only the positive value makes sense chemically, so

[H₃O⁺]_e = 0.0050 + x = 0.0050 + 0.0029 = 0.0079 M

pH = –log(0.0079) = 2.10

Given that HSO₄^– is a pretty strong weak acid, this is a reasonable answer.

 

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