# Why is the second reaction being neglected?

If we want to find the concentration of $\ce{H+}$ ions at equilibrium when a weak conjugate base such as $\ce{C2H3O2-}$ added to water, why do we take the conjugate base reaction for calculation i.e $$\ce{C2H3O2- + H2O \rightarrow C2H3O2H + OH-}$$ Instead of the backward reaction for the weak acid i.e $$\ce{C2H3O2- + H3O+ \rightarrow C2H3O2H + H2O}$$ In spite of the fact that the latter reaction dominates the former since the equilibrium constant of the latter lies far toward the right while that of the former lies far toward the left?

It seems to me that the latter will play a greater part in furnishing $\ce{H+}$ ions than the former does because the former produces very less amount of $\ce{OH-}$ ions to affect the ionic product of water significantly in comparison to the latter. Yet, the latter reaction is neglected while solving the given problem. Where am I going wrong?

• Why don't you save yourself formatting trouble and use the mhchem package? Visit this page to ‎see what I mean. – M.A.R. ಠ_ಠ Feb 10 '15 at 16:08
• How relevant is the second reaction? With other words: What is the concentration of $\ce{H3O+}$ in water before adding the acetate? – Klaus-Dieter Warzecha Feb 10 '15 at 16:22

There is an equilibrium among five species.

$\ce{H2O}$

$\ce{H3O+}$

$\ce{OH-}$

$\ce{CH3COO-}$

$\ce{CH3COOH}$

There are several valid equations you could write for reactions between pairs of these species.

What is important is solving for the concentration of each species.

Unsually it is assumed that water is ~55M, leaving you with 4 unknown concentrations, so you need 4 concentration equations to solve.

$K_a$ gives you one equation.

$K_w$ is a second equation.

$[\ce{CH3COO-}] + [\ce{CH3COOH}] =$ (known constant) is thrid and

$[\ce{CH3COO-}] + [\ce{OH-}] = [\ce{H3O+}]$, ensuring electrical neutrality is fourth.

So both of the chemical equations in the question can happen, but neither is sufficient to describe the actual equilbrium among the five species.

• Thanks for the time, Dave. Forgive me for that I couldn't reply earlier. It's a fine answer and I understand the points you have made here. I have actually already agreed with you on the last statement of your answer when I say that "..yet the latter reaction is neglected while the former is not..". My question is why it is being neglected while solving the particular problem in my book. It must seem logical that the latter reaction must atleast be taken into account, if not being more influential in affecting the ionic product than the former. Answering that, your answer will become complete. – Gaurav Feb 11 '15 at 14:11
• without seeing the book, how can I really answer that? You could try citing the book and hoping I can see it online somehow, or copying that particular section. You haven't explained how mathematically the book is relying on a particular chemical equation in solving the problem. I can't say if the book is approximately correct, completely correct or totally wrong without more information. – DavePhD Feb 11 '15 at 14:31
• Sure. I have added a screenshot of the page to my question of the book where the problem is tackled. Only the former equation is taken into account, and not the latter, as you can see. – Gaurav Feb 11 '15 at 15:01
• Great, very clear now, they are using Kb of aceate instead of Ka of acetic acid, so it makes sense to write the chemical equation the way the book writes it. Basically, you can get the same answer using Ka or Kb, and it is natural to write the chemical equation to correspond to the definition of Ka or definition of Kb, which ever you choose to use. – DavePhD Feb 11 '15 at 15:25
• Okay, your explanation seems very clear to me now. Thanks a lot for the time and effort! – Gaurav Feb 12 '15 at 9:26

It doesn't matter which reaction you take. The short answer as to why is that adding the reaction $\ce{2H2O <-> H3O+ +OH-}$ to the latter reaction gives you the former.

And that water dissociation equilibrium ($K_w$) is assumed to be happening in tandem with any buffer acid-base equilibrium ($K_a$) anyway.