Question

Why can we ignore the contribution of water to the concentrations of H3O+ in the solutions of following acids:

0.0092 M HClO, a weak acid

0.0810 M HCN, a weak acid

0.120 M Fe(H2 O)6 2+ a weak acid, Ka = 1.6 × 10−7

but not the contribution of water to the concentration of OH−?

Answer 1

All the given acids are weak acids. But their contribution to hydronium ion concentration in solution is much more than the contribution by water. So we can ignore the contribution of [H₃O⁺] by water in the solutions of these weak acids.

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For example, consider HClO. Its Ka at 25°C=3.0x10^-8

Consider its dissociation in solution

  

Initial concentration (M) 0.0092 0 0

Change -y +y +y

Equilibrium concentration (M) 0.0092-y y y

Ka==y x y/(0.0092-y)

3.0x10^-8=y²/(0.0092) (ignore y in denominator as Ka is very small so y<<<0.0092 M)

3.0x10^-8 x 0.0092=y^{​​​​​​2}

y^{​​​​​2}=0.0276 x 10^-8 M

y=0.166 x10^{​​​​-4} M=[H₃O⁺]

So we can see that , hence we can ignore the contribution from water

In the solutions of these weak acids, the only contributor of [OH^-] is water, so we can't ignore the contribution of water to the concentration of [OH^-].