Chemistry 112

HClO(aq) + H₂O(l) → ← H₃O⁺(aq) + ClO^–(aq)

K_a = [H₃O⁺]_e[ClO^–]_e                       [HClO]_e = 2.9×10^–8

Initial0.010 0 0

Change–x +x +x

Equilibrium0.010 – x x x

Since K_a is pretty small, x will also be small (we are expecting it to be about 10^–5 to 10^–6). This means the chemistry has done a little bit of algebra for us:

0.010 – x ~ 0.010

x = 1.7×10^–5 = [H₃O⁺]_e

pH = –log[H₃O⁺] = -log(1.7 ×10^–5) = 4.77

 

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