# Concentrations at equilibrium (strong base weak acid)

## Summary:

Why do my concentrations at equilibrium not add up to my initial amount?
So my professor covered acids and bases recently, and this is how he presented the method (please see attached word document). I believe its the same as the I.C.E method, though im not sure. Anyways my problem is perhaps conceptual, but I dont understand why the concentrations at equilibrium dont add up to the amount added in the first place. The first step is to calculate the initial concentrations for the acetic acid at equilibrium by finding alpha. Once you have alpha you can plug it in to the table to find the concentrations. I thought id check that the concentrations at equilibrium added up to 0.1M but they add up to more. Where am I going wrong here?

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etotheipi
Gold Member
2019 Award
I think I understand what might be confusing you but if otherwise please let me know.

With your first ICE table, you have c0 of acetic acid before the reversible reaction begins. Now your reaction equation tells you that 1 mole of acetic acid reacts with 1 mole of water to produce 1 mole of ethanoate ions and 1 mole of hydronium ions. The volume is constant so it is fine to use concentrations from now on.

If the proportion of acetic acid which reacts is $\alpha$, then as is stated your equilibrium concentration of acetic acid is $(1-\alpha)c_{0}$, since $\alpha c_{0}$ of acetic acid has reacted. Now according to the equation, $\alpha c_{0}$ of acetic acid reacts to produce $\alpha c_{0}$ of ethanoate and $\alpha c_{0}$ of hydronium.

You are entirely correct that the final concentrations do not add up to the initial concentration; this is perfectly fine - consider the following reaction:

$A \rightarrow B + C$

The equation tells us the ratio of moles: 1 mole of A reacts to produce 1 mole of B and 1 mole of C. If we start with 10 moles of A, we will end up with 10 moles of B and 10 of C - which adds up to 20 moles!

The important point is that in a chemical reaction, it is not the number of moles (and by extension concentration) which is conserved, since molecules can split up and do all sorts of stuff (perhaps imagine a decomposition reaction where a molecule might split into two equal fragments, we'll end up with twice as many molecules at the end). What is conserved in a chemical reaction is mass.

If you calculate the total mass at the start and end using the $M_{r}$ of each compound then hopefully the mass will have been conserved (!).

• nofluxgiven and Mr-Thirty
Thank you so much, I knew there was a concept that I didnt quite understand. Really appreciate you typing out a thorough response :)

• etotheipi