# How does adding oxalic acid to water change the volume of the solution?

Does $$\pu{87.4 mL}$$ of $$\ce{H2O}$$ and $$\pu{12.6 g}$$ of $$\ce{HOOC-COOH.2H2O}$$ create $$\pu{100 mL}$$ of a solution?

So in our practical class our teacher wanted to make $$\pu{100 mL}$$ $$\pu{1 M}$$ aqueous solution of $$\ce{HOOC-COOH.2H2O}$$ [Oxalic Acid Dihydrate]. So first he measured $$\pu{12.6g}$$ $$\pu{(0.1 mol)}$$ of $$\ce{HOOC-COOH.2H2O}$$, then he measured $$\pu{(100-12.6) mL} = \pu{87.4 mL}$$ of $$\ce{H2O}$$ on a beaker. Then he mixed the $$\pu{12.6g}$$ $$\pu{(0.1 mol)}$$ of solute with the $$\pu{87.4 mL}$$ of solvent.

I have some confusion about this. Because usually for preparing such a solution I would first add $$\pu{12.6 g}$$ of solute on a beaker and then keep adding water until the volume reaches $$\pu{100ml}$$. But I have never seen such a process for creating a solution by measuring a specific volume of solvent and a specific volume of solute and mixing them to get a specific volume of solution. Like: $$a$$ volume of solvent + $$b$$ volume of solute = $$a+b$$ volume of solution.

Can someone explain the theory behind this process?

• Guess your instructor was just having a bad day.. Check this Chem SE post out.. It might help. Aug 10, 2022 at 15:22

There are 2 false assumptions in the proposed procedure:

1. Density of the solute is the same as density of water.
• Most solids are significantly denser than water. Density of oxalic acid dihydrate(OAD) is $$\pu{1.653 g cm-3}$$
2. The total volume does not change when the solute is being dissolved.
• The total volume can shrink or expand. This factor is usually less significant than solute density, but e.g. mixing equal volumes of ethanol and water leads to about 5% volume contraction.

These assumptions may be acceptable if the desired molarity is intended to be just orientational and the volume deviation is acceptable.

The initial total volume of water + OAD, before dissolving, would be: $$V_\mathrm{tot} = \pu{87.4 mL} + \frac{\pu{12.6 g} }{ \pu{1.653 g mL-1}} = 9\pu{5.0 mL}$$

Generally, there is no way to predict the final solution volume (or total volume change) with the given solute mass and solvent volume without knowledge of density of the final solution. If such data is not available, the initial volume can be taken as approximate estimation.

For solution of the given molarity, the standard procedure is proper solute dissolving in such a solvent volume that total volume is somewhat less than final total volume. When dissolved, the volume is then topped to the desired nominal value by the solvent.

Alternatives need to know the final solution density:

• The procedure calculates the needed solvent volume (or mass) for the target nominal volume.
• The procedure calculates the solute mass for given solvent volume, leading to not a well rounded final volume.