# Are aqueous (aq) elements involved in equilibrium?

The reaction was something like this: A (s) + B (l) yields to C (aq) + D (aq)

The question asked: If you added another D, which way would the equilibrium shift to?

I answered that it would have no effect because solids and liquids don't have any effect on the shift of equilibrium, but apparently the correct answer was, it shifts left. I know that would work for gases, but I thought that aqueous means liquid, so that is why I said no effect.

Anyways, I don't understand why it counts. I searched up that aqueous solutions happen when a substance is dissolved in a liquid. So from the term "dissolved", doesn't that mean it is a liquid?

• Aqueous means dissolved in water, not just any liquid. – Ed V Apr 2 at 1:38
• @Ello In your interpretation, the term pH would have no sense, as it is just "liquid" H+ with activity equal 1 and all water solutions would have pH=0. :-) – Poutnik Apr 2 at 2:05
• The three simple states of matter are solid noted by (s), liquid noted by (l) and gas noted by (g). The notation (aq) means the ion or molecule is dissolved in water. So A is a solid, B is a liquid not miscible with water but reacting with A, and the products C/D then dissolve into the water phase, not into liquid B. – MaxW Apr 2 at 9:12
• What is "B (I)", namely capital letter "i" in brackets? Did you mean "B(l)" for a liquid component B? Anyway, try to write down an expression for $K_c$ and see how $[\ce{D(aq)}]$ affects its value. – andselisk Apr 2 at 11:12
• @andselisk I changed it to (l). – Karsten Theis Apr 2 at 11:15

The amount of pure solid or liquid does not affect equilibrium, having chemical activity independent on amount, formally assigned to 1.

If a solvent is in abundance wrt solutes, it is often considered still as approximately pure solvent, e.g water in context of diluted water solutions.

The aqueous (aq) compounds definitely are not pure liquid. They are just dispersed among molecules of water and their chemical activity is approximately proportional to concentration. Therefore, increasing their concentration pushes equilibrium to the other side of the reaction and vice versa.

There is a big touch difference between $$\ce{NaCl(aq)}$$ and $$\ce{NaCl(l)}$$. The former serves for sea bathing/swimming, the latter would make from your body a piece of carbon within seconds.

See e.g. Solubility equilibrium.

I was thinking of a practical example where such an equation could be applied. It is not easy. A solid should react with a liquid, both being insoluble in water. And this reaction should produce two substances both soluble in water. I don't see many possibilities for this reaction. A may be a metal (maybe iron $$\ce{Fe},$$ or even gold $$\ce{Au}),$$ and $$\ce{B}$$ some bromine $$\ce{Br2}.$$ So the reactions could look like:

\begin{align} \ce{Fe + Br_2 &-> Fe^{2+} + 2 Br^-}\\ \ce{2 Au + 3 Br_2 &-> 2 Au^{3+} + 6 Br^-} \end{align}

Would you believe that adding some $$\ce{NaBr}$$ in a $$\ce{FeBr2}$$ or $$\ce{AuCl3}$$ solution will produce more metal and more $$\ce{Br2}?$$ Is my example poorly chosen? I would be glad to read your reaction and your interpretation of such an example.

• While I think this is an ingenious answer which I happily upvoted, I would recommend not to ask direct non-rhetorical questions. Forums are great for that, but the Q&A sites are not: it's the wrong format. Here, all supplementary questions and discussion with OP should be sorted out in the comment section, and the answer should be complete and not open-ended. – andselisk Apr 2 at 11:19

The more intuitive reason that substances labelled with (aq) are taken into account for shift in the state of equilibrium:

Let us consider activity as concentration of the substance for the sake of this question.

The concentration of solids and liquids is taken as unity while studying equilibrium because they are incompressible in most cases. So if you add say 100mL of NaCl (l) to 1000mL of NaCl (l) present already, then that 1000mL still takes up the same volume in the reaction mixture while the new 100mL takes up new space(volume). Also note that 100mL and 1000mL are comparable here.

On the other hand substances labelled as (aq) are taken in small quantities (for example: 5gm) and spread throughout the vast quantity of water (100mL - 2L). Thus if you add more of the former substance it will enter the same water and the population of this substance would become more in the same volume of water. (Eg. 5g in 1000L water is 0.005 g/L ratio to which if again 5g of that substance is added it becomes 0.01 g/L ratio)

Hence in case of the liquid the concentration remains same and for the substance it increases.

Shift in the equilibrium is shown by reaction quotient

Qc =[D] (aq) [C] (aq) / [A] (s) [B] (l)

As explained above solid and liquids get concentration unity while aqueous substance has variability. Thus [A]=[B]=1.

Therefore reaction quotient (which is a projection of shift in the equilibrium) takes the expression

Qc =[D] (aq) * [C] (aq)