# Why does dilution reduce the concentration of hydronium and hydroxide ions?

So my teacher gave us the definition of dilution:

Mixing of an acid/base with $\ce{H2O}$ to reduce the concentration of $\ce{H+/OH-}$ ions per unit volume.

But doesn't mixing an acid/base to water increase the concentration of $\ce{H+/OH-}$ ions?

I don't understand this please explain.

• Compared to the acid/base alone, it decreases the concentration. – f'' Aug 3 '16 at 7:29
• How? I dont understand – MartianCactus Aug 3 '16 at 7:41

He is talking about dissociated H+/OH-. Consider, for example, a NaOH (strong base) aqueous solution: You have a dissociation of Na+ and OH- ions. Of course, the OH- ions give the solution its basicity.

By increasing the dilution of the solution, the concentration of dissociated OH- anions decreases, and thus the basicity of the solution decreases.

• so we are adding in more water? – MartianCactus Aug 3 '16 at 8:37
• Yes, because you could consider (although it's not 100% correct, but let's consider the concept) as the OH- and H+ ions of water were not dissociated, but instead they were completely aggregated in the H2O molecule. The H+ or OH- ions of the added acid or base, instead, are dissociated. By increasing the dilution, you are decreasing the concentration of the ions that come from the acid/base, and so decreasing the acidity/basicity. – The_Vinz Aug 3 '16 at 8:50

Let’s take an example base which can be said to be fully dissociated in water, e.g. $\ce{NaOH}$. You take $40~\mathrm{g}$ of the pure compound and dissolve it in $0.5~\mathrm{l}$ of water. The standard way to calculate the concentration of sodium ion is:

$$c = \frac nV = \frac m{MV} = \frac{40~\mathrm{g}}{40~\mathrm{g\cdot mol^{-1}} \times 0.5~\mathrm{l}} = 2~\mathrm{\frac {mol}l}\tag{1}$$

This builds on the fact that the amount of sodium ions is identical to the amount of $\ce{NaOH}$ added. The same thing can be done for hydroxide ions; within experimental error the concentration of hydroxide ions is also $2~\mathrm{mol\cdot l^{-1}}$ as calculated in equation $(1)$.

Now say you add another $0.5~\mathrm{l}$ to the solution you generated above. Obviously, you did not add any additional sodium ions, so their amount will remain the same. The new concentration can be calculated as shown in $(2)$:

$$c' = \frac n{V'} = \frac m{MV'} = \frac{40~\mathrm{g}}{40~\mathrm{g\cdot mol^{-1}} \times 1.0~\mathrm{l}} = 1~\mathrm{\frac {mol}l}\tag{2}$$

Thus, in the resulting solution, you are left with half the original concentration of sodium ions; similarly for hydroxide ions at first approximation. The concentration decreased by adding water.

Why did I always include a caveat with the hydroxide ions? Well, these are also determined by the autoionisation of water, defined by its equilibrium constant $K_\mathrm{w}$ of equation $(3)$ as shown in $(4)$.

$$\begin{gather}\ce{H2O <--> H+ + OH-}\tag{3}\\[0.6em] K_\mathrm{w} = [\ce{H+}][\ce{OH-}] = 10^{-14}~\mathrm{\frac{mol^2}{l^2}}\tag{4}\end{gather}$$

Since the concentration of added hydroxide is much larger than that present in pure water ($1 \gg 10^{-7}$), we can ignore the additional concentration of hydroxide introduced by the equilibrium. However, we can calculate a new concentration of $\ce{H+}$, see equation $(5)$:

$$[\ce{H+}] = \frac{K_\mathrm{w}}{[\ce{OH-}]} = \frac{10^{-14}~\mathrm{mol^2 \cdot l^{-2}}}{2~\mathrm{mol \cdot l^{-1}}} = 5 \cdot 10^{-15}~\mathrm{\frac{mol}l}\tag{5}$$

Thus, after dilution, the new concentration of protons is:

$$[\ce{H+}]' = \frac{K_\mathrm{w}}{[\ce{OH-}]'} = \frac{10^{-14}~\mathrm{mol^2 \cdot l^{-2}}}{1~\mathrm{mol \cdot l^{-1}}} = 1 \cdot 10^{-14}~\mathrm{\frac{mol}l}\tag{6}$$

Note how this concentration increased; this is reflected in the change of the solution’s $\mathrm{pH}$ value from $14.3$ to $14$.

Suppose there is one mole of $\ce{NaOH}$ in 1 liter of $\ce{H2O}$ so it is $1~\mathrm{M}\ \ce{NaOH}$. Now you dilute it to 2 liters of solution you have $0.5~\mathrm{M}$ of $\ce{OH-}$ in the solution. The added $\ce{H2O}$ does not disassociate.