Why does concentration decrease when water is added

Water, Acids, and Bases

The Acid-Base Chemistry of Water

The chemistry of aqueous solutions is dominated by the equilibrium between neutral water molecules and the ions they form.

2 H2O(l) H3O+(aq) + OH-(aq)

Strict application of the rules for writing equilibrium constant expressions to this reaction produces the following result.

This is a legitimate equilibrium constant expression, but it fails to take into account the enormous difference between the concentrations of neutral H2O molecules and H3O+ and OH- ions at equilibrium.

Measurements of the ability of water to conduct an electric current suggest that pure water at 25oC contains 1.0 x 10-7 moles per liter of each of these ions.

[H3O+] = [OH-] = 1.0 x 10-7 M

At the same temperature, the concentration of neutral H2O molecules is 55.35 molar.

The ratio of the concentration of the H+ (or OH-) ion to the concentration of the neutral H2O molecules is therefore 1.8 x 10-9.

In other words, only about 2 parts per billion (ppb) of the water molecules dissociate into ions at room temperature.

The equilibrium concentration of H2O molecules is so much larger than the concentrations of the H3O+ and OH- ions that it is effectively constant. We therefore build the [H2O] term into the equilibrium constant for the reaction and thereby greatly simplify equilibrium calculations. We start by rearranging the equilibrium constant expression for the dissociation of water to give the following equation.

[H3O+][OH-] = Kc x [H2O]2

We then replace the term on the right side of this equation with a constant known as the waterdissociation equilibrium constant, Kw.

[H3O+][OH-] = Kw

In pure water, at 25C, the [H3O+] and [OH-] ion concentrations are 1.0 x 10-7 M. The value of Kw at 25C is therefore 1.0 x 10-14.

Although Kw is defined in terms of the dissociation of water, this equilibrium constant expression is equally valid for solutions of acids and bases dissolved in water. Regardless of the source of the H3O+ and OH- ions in water, the product of the concentrations of these ions at equilibrium at 25C is always 1.0 x 10-14.

Strong Acids and the H3O+ and OH- Ion Concentrations

Suppose we add enough strong acid to a beaker of water to raise the H3O+ ion concentration to 0.010 M. According to LeChatelier's principle, this should drive the equilibrium between water and its ions to the left, reducing the number of H3O+ and OH- ions in the solution.

2 H2O(l) H3O+(aq) + OH-(aq)

Because there are so many H3O+ ions in this solution, the change in the concentration of this ion is too small to notice. When the system returns to equilibrium, the H3O+ ion concentration is still about 0.010 M. Furthermore, when the reaction returns to equilibrium, the product of the H3O+ and OH- ion concentrations is once again equal to Kw.

[H3O+][OH-] = 1 x 10-14

The solution therefore comes back to equilibrium when the dissociation of water is so small that the OH- ion concentration is only 1.0 x 10-12 M.

Adding an acid to water therefore has an effect on the concentration of both the H3O+ and OH- ions. Because it is a source of this ion, adding an acid to water increases the concentration of the H3O+ ion. Adding an acid to water, however, decreases the extent to which water dissociates. It therefore leads to a significant decrease in the concentration of the OH- ion.

As might be expected, the opposite effect is observed when a base is added to water. Because we are adding a base, the OH- ion concentration increases. Once the system returns to equilibrium, the product of the H3O+ and OH- ion concentrations is once again equal to Kw. The only way this can be achieved, of course, is by decreasing the concentration of the H3O+ ion.

The reaction $$\ce{NaCl~(s) <=>[\ce{H2O}] Na+~(aq) + Cl^{-}~(aq)}$$ is at equilibrium when approximately $359~\mathrm{g/L}$ sodium chloride are dissolved in water, i.e. the solution is saturated, or $c(\ce{Na+}) = c(\ce{Cl-}) = c_\mathrm{max}$.

If less $\ce{NaCl}$ is dissolved, it is only one phase, which is in equilibrium with itself and all solid salt is dissolved, therefore the equilibrium would be entirely on the right side. At that point adding water would not change the equilibrium at all, it would only dilute the whole system.

When the solution is saturated, there is still solid salt present, the equilibrium is not entirely on the right side. Adding water will disturb the equilibrium, i.e. $c_\mathrm{present}<c_\mathrm{max}$. Therefore more of the excess $\ce{NaCl}$ will dissolve, until the maximum concentration is reached again. If the solution is still saturated after that, nothing will have changed in your solution except for the volume.
If your solution is not saturated anymore, the equilibrium is entirely on the right side and the concentrations will decrease.

Updated July 21, 2017

By Serm Murmson

In a typical titration, you add a known quantity of a reagent called a titrant to an analyte. The analyte is a solution of unknown concentration. As you slowly add the titrant, you can monitor for signs that a reaction is taking place. Water is necessary to create the solutions in titrations. Additionally, if you add water to a solution, you change the concentration of the solution. You must incorporate these changes into your calculations.

Adding water to a titrant or analyte will change the concentration of that solution. Each solution has a molarity, which is equal to the number of moles of a solvent per liter of solution. When you add water to a solution, the number of moles of the solvent stays the same while the volume increases. Therefore, the molarity decreases; the solution is diluted.

When you add water to the analyte, you dilute a solution of unknown molarity. This dilution ultimately does not affect the experimental results. The concentration of the analyte is still unknown. As long as your volume measurements are accurate, you can calculate the moles of the unknown compound after the titration is complete.

When you add water to the titrant, you dilute a solution of known molarity. This is important to factor into your calculations at the end of the titration; you must know the number of moles of the titrant used in the titration. As long as you incorporate the added water into your calculations, your results should be accurate. Also, because you dilute the titrant, it will take a larger amount of titrant to cause a change in the analyte. Therefore, the entire titration process will take longer.

Most titrations depend on precise pH measurements. Water has pH of seven, which is neutral. When you add it to an acid or base, it dilutes that solution and brings the pH closer to seven. As long as you account for this dilution in your titration calculations, the addition of water should not cause errors in your results.