The following graph shows $\ce{Ka}$ vs ${V}$ (Volume) at constant temperature, as ethanoic acid is diluted:

enter image description here

However, I thought it would instead produce the following graph (see below).


As ${V}$ increases, I would think all of $\ce{[H3O+]}$,$\ce{[A-]}$ and $\ce{[HA]}$ would decrease similarly. As there are two concentrations decreasing in the numerator, and only one decreasing in the denominator, this leads me to believe I can write this as $\ce{Ka ∝ [X]}$, where $\ce{[X]}$ decreases as ${V}$ increases (i.e. that I could cancel out one of the concentration in both the numerator and denominator, leaving only $1$ concentration in the numerator; noting that during ionisation $1$ mole of $\ce{[HA]}$ turns into $1$ mole of $\ce{[H3O+]}$and $1$ mole of $\ce{[-A]}$)

Giving the following graph:

enter image description here

Why is the top and not the bottom graph produced?



The above equation defines the $K_\mathrm{a}$, or acid dissociation constant, of an acid. The reason it is even a value worth measuring is because it is an intrinsic property of the acid, i.e. it's value does not change from sample to sample (temperature aside).

While it is true that as a solution of acid becomes more dilute, $\ce{[A-]}$, $\ce{[HA]}$ and $\ce{[H3O+]}$ all decrease, they do not all decrease by the same amount. Consider the effect of adding water to the following equilibrium:

$$\ce{HA + H2O <=>A- +H3O+}$$

The number of $\ce{HA}$ molecules will decrease, and the number of $\ce{A-}$ and $\ce{H3O+}$ molecules will increase—a conclusion we arrive at via Le Châtelier's principle. Separating the volume of water from the $K_\mathrm{a}$ equation, we get:

$$K_\mathrm{a}=\ce{\frac{mol(A-)\cdot mol(H3O+)}{mol(HA)}}\cdot\frac{1}{V}$$

What we are left with is two larger numbers over a smaller and a larger number. While it's not obvious that these numbers should change to give precisely the same quotient, I can assure you they do.

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    $\begingroup$ Right: the key to realize is that although for a theoretical instantaneous dilution all of the concentrations would be diluted by the same amount, after the dilution you're no longer (necessarily) at equilibrium. The compounds will then react to reestablish equilibrium, which means their concentrations will change until their equilibrium ratio is equal to the $\ce{K_a}$. $\endgroup$ – R.M. Jan 15 '17 at 16:44

Ka of weak acid is an equilibrium constant and equilibrium constant depends upon temperature and its value increases with increase in temperature. During dilution Qa of weak acid will change and its value will decrease. When Qa < Ka( as Ka will remain constant) reaction will not remain at equilibrium and it shift towards forward direction to attain equilibrium.

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    $\begingroup$ This is the better answer. Changing the volume, changes Q. Then the concenatrations shift to bring Q back to K. $\endgroup$ – Ben Norris Jan 15 '17 at 14:41

The reaction of an acid in water is of the form

$$\ce{HA + H2O <=> H3O^+ + A^-}$$

When two of the species in solution come close enough to interact, then there is a certain probability that they will react to form products. The probability mainly depends on the type of the involved species and much less on the surrounding solution.

However, the probability that two species meet is proportional to their concentration (not volume). Therefore we can formulate the rate law

$$\frac{d[\ce{HA}]}{dt} = -k_1 [\ce{HA}] [\ce{H2O}] \tag{1}$$

for the forward reaction

$$\ce{HA + H2O -> H3O^+ + A^-}$$

and the rate law

$$\frac{d[\ce{A-}]}{dt} = -k_2 [\ce{H3O+}] [\ce{A-}] \tag{2}$$

for the backward reaction

$$\ce{H3O+ + A- -> HA + H2O}.$$

The rate constants $k_1$ and $k_2$ hereby reflect the said reaction probabilities.

In the thermodynamic equilibrium the reaction rates for forward and backward reaction are the same. By setting equations $(1)$ and $(2)$ equal we obtain

$$k'_a = \frac{k_1}{k_2} = \frac{[\ce{H3O+}] [\ce{A-}]}{[\ce{HA}] [\ce{H2O}]} \tag{3}$$

Since the concentration of water in a dilute aqueous solution can be considered constant, we normally set

$$k'_a [\ce{H2O}] = k_a \tag{4}$$

to get the familiar equation

$$k_a = \frac{[\ce{H3O+}] [\ce{A-}]}{[\ce{HA}]} \tag{5}$$

Neither equation $(3)$ nor equation $(5)$ depends on the volume $V$ of the solution.

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