The question title is a bit odd, I couldn't find a plainer way of saying it.

I'm given the concentration and conductivity of a solution of a particular acid, let's just call it $\ce{AH}$. I then have to deduce that acid's acidity constant.

My first instinct is to write out the expression for the solution's conductivity... but then I realize I don't know how to do that. I have the coefficients for all the relevant ions (I'm not sure what the technical term is, but essentially I mean the lambdas in the second equation here), in particular, the ions $\ce{H_3O^+}$, $\ce{OH^-}$ and the conjugate base $\ce{A^-}$.

So if I had the concentrations of those ions in the solution, I could solve the problem! But hang on... I have no idea how to figure them out. I know the concentration of the acid in the solution. Suppose the acid has just been added to the water. Then it won't have started to disassociate yet, so the water will be exactly as it was before we added the acid: $[\ce{H_3O^+}] = [\ce{OH^-}] = 10^{-7}$. But what if originally there was a lot more acid, but it's had time to react with the water? Then the water will have been filled with an amount of $\ce{A^-}$ dependent on how long ago the acid was dissolved in it.

Do I assume the acid is at equilibrium? The question doesn't specify that.

  • $\begingroup$ A solution of an acid usually refers to it's aqueous solution -- also written as $\ce{AH(aq)}$. And as described in Wikipedia - a solution is a homogeneous mixture. So yes, the acid will be dissolved, i.e. in equilibrium. $\endgroup$ – Alex Apr 1 '13 at 0:04
  • $\begingroup$ That's not strictly true for acids and bases though. What makes an acid an acid is the fact that it gives off protons, especially in water, which react with water molecules to form $H_{3}O^{+}$. So what you got when you dissolve an acid $AH$ in water is really a mixture of $AH$, $H_{2}O$, $A^{-}$ and $H_{3}O^{+}$. The concentrations of $AH$ and $A^{-}$ relative to each other at a given concentration in water is what the specific acidity constant of an acid is, and that's what the OP is supposed to find out. $\endgroup$ – Tanith Rosenbaum Apr 1 '13 at 0:54

The acidity constant is defined as the equilibrium constant for the reaction:

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

$$K_\text{a} = \dfrac{[\ce{H^+}][\ce{A^-}]}{[\ce{HA}]}$$

Note: $\ce{H3O+}$ can be considered equivalent to $\ce{H+}$

Since you have to find an equilibrium constant, (and no further info is given) you can safely assume the solution to be in equilibrium.

Plug in the various conductivities in the second equation (Kohlrausch's Law) you mentioned getting the limiting conductivity ($\lambda^o$) of the acid. This is the conductivity of the acid if it had completely dissociated into its constituent ions.

EDIT: To find the limiting conductivity of the acid using the equation, you must have the values of $\lambda^o_{\ce{H^+}}$ and $\lambda^o_{\ce{A^-}}$.

$$\lambda^o_{\ce{HA}} = \lambda^o_{\ce{H+}} + \lambda^o_{\ce{A-}}$$

The degree of dissociation of the acid $\alpha$ (The number of acid molecules which have dissociated at equilibrium per each acid molecule present originally -- it is fractional for weak acids and bases) would be the ratio of the measured conductivity of the solution ($\lambda$, which is given) to the limiting conductivity ($\lambda^o$).

$$\alpha = \dfrac{\lambda_{\ce{HA}}}{\lambda^o_{\ce{HA}}}$$

This alpha would be the same as the ratio of concentration of dissociated acid molecules at equilibrium (since each dissociated acid molecule gives one $\ce{H^+}$ ion, this is equal to the number of $\ce{H^+}$ ions at equilibrium) to the initial concentration of acid taken (which you have mentioned as given).

Therefore, $$\dfrac{\lambda_{\ce{HA}}}{\lambda^o_{\ce{HA}}} = \dfrac{[\ce{H^+}]}{[\ce{HA}]_\text{initial}}$$

You can find $[\ce{H^+}]$ at equilibrium using this equation and then find the other concentrations using that.

Note that all this assumes that the acid is monoprotic (i.e. there is only one proton released per acid molecule). You would have to change all the equations accordingly in case more than one proton is released.


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