Calculation of Degree of Dissociation from Ostwald's Dilution Law

Question

A "weak" electrolyte, $\ce{A+B-}$, ionizes in solution as:

$$\ce{AB <=> A+ + B-}\tag{1}$$

$$ K_d =\dfrac{\ce{[A+][B-]}}{\ce{[AB]}}=\dfrac{(\alpha c_{0})(\alpha c_{0})}{(1-\alpha )c_{0}}=\dfrac{\alpha ^{2}}{1-\alpha }\cdot c_{0}\tag{2}$$

For a weak electrolyte the Degree of Dissociation, $\alpha$, is less than 1 and can be calculated using two different equations.

• Equation (2) can be rearranged to a quadratic equation which could be solved exactly with the standard quadratic formula.

$$ c_0 \cdot \alpha^2 + K_d\cdot \alpha -K_d = 0\tag{3}$$

$$\alpha = \dfrac{-K_d + \sqrt{K_d^2 - 4\cdot c_0 \cdot (-K_d) }}{2\cdot c_0} = \dfrac{-K_d + \sqrt{K_d^2 + 4\cdot c_0 \cdot K_d }}{2\cdot c_0}\tag{4}$$

• If we assume that there is little dissociation and that $\ce{[AB] \approx c_0}$ then $1- \alpha \approx 1$ and we obtain an approximate answer that is easier to calcualte:

$$ \alpha' \approx \sqrt{\dfrac{K_d}{c_{0}}}\tag{5}$$

If equation (5) does not give the true value of the Degree of Dissociation. How do you know when it is good enough?

To make it clearer, say the approximate value from formula (5) is 0.15. That is not the true value of the degree of dissociation, but would the approximate solution be good enough?

Answer 1

The OP asked a simple question to which I gave a complex answer. My daughter called such answers "Dad answers".

The OP essentially asked "If the dilute solution approximation for $\alpha$ is used when calculating, how do you know how good the approximation is?"

In his answer user porphyrin focused on the pertinent idea.

Dilute Solution Approximation

Now if we assume that $(1-\alpha) \approx 1$ meaning that very little of the compound dissociated, then we get the standard approximation, $\alpha'$:

$$ \alpha^{'} \approx \sqrt{\dfrac{K_d}{c_{0}}}\tag{1}$$

Exact Solution

The equation below is an exact solution even though it doesn't allow $\alpha$ to be calculated directly. You would have to guess at $\alpha$ values and iterate until you found a solution.

$$\dfrac{\alpha}{\sqrt{1 - \alpha}} = \sqrt{\dfrac{K_d}{c_{0}}}\tag{2}$$

Estimated % Error

Now let's calculate an estimated % error as:

$$\pu{Est \% Error} = 100 \times \dfrac{\dfrac{\alpha'}{\sqrt{1 - \alpha'}} - \alpha'}{\alpha'}\tag{3}$$

Quadratic Solution

The exact value of $\alpha$ can be calculated directly using a quadratic equation. The solution is given below.

$$\alpha = - \dfrac{K_d}{2\cdot c_0} + \sqrt{\left(\dfrac{K_d}{2\cdot c_0}\right)^2 + \dfrac{K_d}{c_0}}\tag{4}$$

Data Table

\begin{array}{|c|c|c|c|c|} \hline (\alpha^{'})^2=K_d/c_0 & \alpha (quadratic) & approx & true\ \% \ error & est.\ \% \ error \\ \hline 1.00000000 & 0.61803399 & 1.00000000 & 61.80339887 & \\ \hline 0.81000000 & 0.58192705 & 0.90000000 & 54.65856100 & 216.22776602 \\ \hline 0.64000000 & 0.54162637 & 0.80000000 & 47.70329614 & 123.60679775 \\ \hline 0.49000000 & 0.49663670 & 0.70000000 & 40.94810050 & 82.57418584 \\ \hline 0.36000000 & 0.44641839 & 0.60000000 & 34.40306509 & 58.11388301 \\ \hline 0.25000000 & 0.39038820 & 0.50000000 & 28.07764064 & 41.42135624 \\ \hline 0.16000000 & 0.32792156 & 0.40000000 & 21.98039027 & 29.09944487 \\ \hline 0.09000000 & 0.25835623 & 0.30000000 & 16.11874208 & 19.52286093 \\ \hline 0.04000000 & 0.18099751 & 0.20000000 & 10.49875621 & 11.80339887 \\ \hline 0.01000000 & 0.09512492 & 0.10000000 & 5.12492197 & 5.40925534 \\ \hline 0.00810000 & 0.08604108 & 0.09000000 & 4.60119879 & 4.82848367 \\ \hline 0.00640000 & 0.07686397 & 0.08000000 & 4.07996803 & 4.25720703 \\ \hline 0.00490000 & 0.06759286 & 0.07000000 & 3.56123125 & 3.69516947 \\ \hline 0.00360000 & 0.05822699 & 0.06000000 & 3.04498988 & 3.14212463 \\ \hline 0.00250000 & 0.04876562 & 0.05000000 & 2.53124512 & 2.59783521 \\ \hline 0.00160000 & 0.03920800 & 0.04000000 & 2.01999800 & 2.06207262 \\ \hline 0.00090000 & 0.02955337 & 0.03000000 & 1.51124937 & 1.53461651 \\ \hline 0.00040000 & 0.01980100 & 0.02000000 & 1.00499988 & 1.01525446 \\ \hline 0.00010000 & 0.00995012 & 0.01000000 & 0.50124999 & 0.50378153 \\ \hline 0.000001 & 0.0009995 & 0.00100000 & 0.05001250 & 0.05003753 \\ \hline 0.00000001 & 9.9995E-05 & 0.00010000 & 0.00500012 & 0.00500038 \\ \hline 1E-10 & 9.99995E-06 & 0.00001000 & 0.00050000 & 0.00050000 \\ \hline \end{array}

So the Est % Error from equation (3) works quite well.

• When the Est % Error is 5% or less, $\alpha' \le 0.1$, the Est % Error is quite good.

• So for the approximate equation when $\alpha^{'} = 0.1$ you get 1 significant figure, 0.01 gets two significant figures and so on. The improvement doesn't continue forever of course - this is chemistry not math. To get three significant figures or more you'd have to be a fastidious experimenter.

• For the approximate equation when $\alpha' = 0.7$ the estimated error is 82.6% which is much too large, but an 82.6% error still tells you that the approximation $\alpha^{'}$ is no good.

• To make user M. Farooq's point in another way, by the time you calculate the approximation from equation (1) and the Est % Error from equation (3), you've pretty much done enough work to calculate the exact answer using equation (4). So why bother with the approximation?

---

EDIT A comment by user Buck Thorn "I never thought about the usefulness of the estimate to bracket the true value" made me think.

It turns out that we can bracket the true value $\alpha$ using a simple calculation of the approximate value $\alpha'$:

$$\alpha'\cdot(1-\alpha') \lt \alpha < \alpha'\tag{5}$$

\begin{array}{|c|c|c|c|} \hline (\alpha^{'})^2=K_d/c_0 & \alpha (quadratic) & \alpha'\cdot(1-\alpha') & \alpha' \\ \hline 1.0000000000 & 0.6180339887 & 0.0000000000 & 1.0000000000 \\ \hline 0.8100000000 & 0.5819270490 & 0.0900000000 & 0.9000000000 \\ \hline 0.6400000000 & 0.5416263691 & 0.1600000000 & 0.8000000000 \\ \hline 0.4900000000 & 0.4966367035 & 0.2100000000 & 0.7000000000 \\ \hline 0.3600000000 & 0.4464183905 & 0.2400000000 & 0.6000000000 \\ \hline 0.2500000000 & 0.3903882032 & 0.2500000000 & 0.5000000000 \\ \hline 0.1600000000 & 0.3279215611 & 0.2400000000 & 0.4000000000 \\ \hline 0.0900000000 & 0.2583562262 & 0.2100000000 & 0.3000000000 \\ \hline 0.0400000000 & 0.1809975124 & 0.1600000000 & 0.2000000000 \\ \hline 0.0100000000 & 0.0951249220 & 0.0900000000 & 0.1000000000 \\ \hline 0.0081000000 & 0.0860410789 & 0.0819000000 & 0.0900000000 \\ \hline 0.0064000000 & 0.0768639744 & 0.0736000000 & 0.0800000000 \\ \hline 0.0049000000 & 0.0675928619 & 0.0651000000 & 0.0700000000 \\ \hline 0.0036000000 & 0.0582269939 & 0.0564000000 & 0.0600000000 \\ \hline 0.0025000000 & 0.0487656226 & 0.0475000000 & 0.0500000000 \\ \hline 0.0016000000 & 0.0392079992 & 0.0384000000 & 0.0400000000 \\ \hline 0.0009000000 & 0.0295533748 & 0.0291000000 & 0.0300000000 \\ \hline 0.0004000000 & 0.0198010000 & 0.0196000000 & 0.0200000000 \\ \hline 1.000000E-04 & 0.0099501250 & 0.0099000000 & 0.0100000000 \\ \hline 1.000000E-06 & 0.0009995001 & 0.0009990000 & 0.0010000000 \\ \hline 1.000000E-08 & 0.0000999950 & 0.0000999900 & 0.0001000000 \\ \hline 1.000000E-10 & 9.999950E-06 & 9.999900E-06 & 0.0000100000 \\ \hline \end{array}

Answer 2

As an alternative and simple estimation, assume $\alpha \lt\lt 1$ and expand $\displaystyle \frac{K_d}{c_0}=\frac{\alpha^2}{1-\alpha}$ to give $\displaystyle \frac{K_d}{c_0}=\alpha^2(1+\alpha+\alpha^2+\alpha^3+\cdots)$ and compare the result to your approximation by adding correction terms.

Answer 3

Further elaboration to the answer. I recall teaching this experiment more than a decade ago (slightly rusty now). I think the approximation was is not mathematical (A) or (B), as one has to solve the entire quadratic equation by all means.

$$ \alpha = \sqrt{\dfrac{K_d\cdot (1-\alpha )}{c}}\tag{A}$$

and

$$ \alpha^{'} \approx \sqrt{\dfrac{K_d}{c}}\tag{B}$$

As one can see in equation (A) and (B), K$_d$ has a very strong dependence on concentration of acetic acid taken for measurement. Certainly, during the experiment we cannot take an infinitely dilute weak acid solution, so we do the experiment with sufficiently dilute acid but of finite concentration. The fundamental or key issue is the way $\alpha$ is defined. It is given by

$$ \alpha = \frac{\Lambda}{\Lambda_0}\tag{C}$$

where the numerator indicates the conductivity at concentration $c$ and the denominator shows the conductivity at zero concentration.

The accuracy of the dissociation constant is dependent on this measurement and the validity of the equation (C). This expression is not valid for strong electrolytes (hence strong bases and acids are not used for dissociation constant experiments)

So apparently, the student is confused why only weak electrolytes can be measured by conductivity: The reason is the validity of equation (C).

He also assumes that only conductivity experiment is used to determine the dissociation constant of an acid and base. Since we are determining Kd from a conductivity measurement, how do we know beforehand that the acid/base is weak. This circular argument is not true at all. The weakness of an acid or a base in water is easily determined by a simple pH measurement. For example, pH of 0.01 M acetic acid is higher than pH of 0.01 M HCl, which indicates that acetic acid is not fully dissociated. Conductivity is not the only way to determine the dissociation constants, it is one of the possible ways.

Answer 4

I rearranged the answer and moved the equations around. However I kept the original numbering so that the comments make sense.

Consider a binary electrolyte AB which dissociates reversibly into A+ and B− ions with a nominal concentration of $c_0$. Ostwald noted that the law of mass action can be applied to such systems as dissociating electrolytes. The equilibrium state is represented by the equation:

$$\ce{AB <=> A+ + B^-}\tag{1}$$

If $\alpha$ is the fraction of dissociated electrolyte, then $\alpha \cdot c_0$ is the concentration of each ionic species. $(1-\alpha)$ must, therefore be the fraction of undissociated electrolyte, and $(1- \alpha )\cdot c_0$ the concentration of same. The dissociation constant may therefore be given as

$$ K_d =\dfrac{\ce{[A+][B-]}}{\ce{[AB]}}=\dfrac{(\alpha c_{0})(\alpha c_{0})}{(1-\alpha )c_{0}}=\dfrac{\alpha ^{2}}{1-\alpha }\cdot c_{0}\tag{2}$$

So far I've copied the Wikipedia article on the Law of dilution almost exactly. Now I'll depart that derivation.

From equation (2) let's start with: $$ K_d = \dfrac{\alpha ^{2}}{1-\alpha }\cdot c_{0}\tag{3}$$

Dilute Solution Approximation

now if we assume that $(1-\alpha) \approx 1$ meaning that very little of the compound dissociated, then we get the standard approximation, $\alpha'$:

$$ \alpha^{'} \approx \sqrt{\dfrac{K_d}{c_{0}}}\tag{6}$$

Exact Solution - Odd Form

Now let's play around with Equation (3) to form an exact solution in an odd sort of form.

$$ \alpha^2 = \dfrac{K_d\cdot (1-\alpha )}{c_{0}}\tag{4}$$

and solving for the square root:

$$ \alpha = \sqrt{\dfrac{K_d}{c_{0}}\cdot (1-\alpha )}\tag{5}$$

Now Equation (5) is in an odd form, but it is exact. Knowing $K_d/c_0$ one could guess at values of $\alpha$ until a value was found that would satisfy the equation.

Remember that we're looking for values of $K_c/c_0 << 1$. Comparing equations (6) and (5) it is obvious to the casual observer that equation (6) must always give a value that is larger than equation (5). Hence if $ 1 \gg \alpha^{'}$ then $ 1 \gg \alpha^{'} \gt \alpha$

Exact Solution - Exact Quadratic Solution

Equation(5) is in the wrong form to directly solve for $\alpha$ of course. So let's go back to equation (3) and rearrange to a reduced quadratic equation which could be solved with the standard quadratic formula.

$$ \alpha^2 +\dfrac{K_d}{c_0}\cdot \alpha -\dfrac{K_d}{c_0}= 0\tag{7}$$

$$\alpha = - \dfrac{K_d}{2\cdot c_0} + \sqrt{\left(\dfrac{K_d}{2\cdot c_0}\right)^2 + \dfrac{K_d}{c_0}}\tag{8}$$

Data Table

Now for various $K_d/c_0$ values let's calculate the approximate solution, the exact solution, and the % Error.

\begin{array}{|c|c|c|c|} \hline \dfrac{K_d}{c_0} & Approximation & Quadratic & \% Error \\ & Equation (6) & Equation (8) & \\ \hline 1.00000000 & 1.00000000 & 0.618034 & 61.803399 \\ \hline 0.81000000 & 0.90000000 & 0.581927 & 54.658561 \\ \hline 0.64000000 & 0.80000000 & 0.541626 & 47.703296 \\ \hline 0.49000000 & 0.70000000 & 0.496637 & 40.948101 \\ \hline 0.36000000 & 0.60000000 & 0.446418 & 34.403065 \\ \hline 0.25000000 & 0.50000000 & 0.390388 & 28.077641 \\ \hline 0.16000000 & 0.40000000 & 0.327922 & 21.980390 \\ \hline 0.09000000 & 0.30000000 & 0.258356 & 16.118742 \\ \hline 0.04000000 & 0.20000000 & 0.180998 & 10.498756 \\ \hline 0.02647076 & 0.16269836 & 0.150000 & 8.465257 \\ \hline 0.02250000 & 0.15000000 & 0.139171 & 7.780856 \\ \hline 0.01000000 & 0.10000000 & 0.095125 & 5.124922 \\ \hline 0.00010000 & 0.01000000 & 0.009950125 & 0.501249992 \\ \hline 1.000000E-06 & 0.00100000 & 9.995001E-04 & 5.001250E-02 \\ \hline 1.000000E-08 & 0.00010000 & 9.999500E-05 & 5.000125E-03 \\ \hline 1.000000E-10 & 0.00001000 & 9.999950E-06 & 5.000013E-04 \\ \hline 1.000000E-12 & 0.00000100 & 9.999995E-07 & 5.000001E-05\\ \hline \end{array}

(1) So for this equation when the approximate amount of dissociation is 0.1 you get 1 significant figure, 0.01 gets two significant figures and so on. The improvement doesn't continue forever of course this is chemistry not math. To get three significant figures you'd have to be a fastidious experimenter.

(2) Notice that the exact quadratic solution is always lower than the approximate solution.

(3) The OP asked how the exact solution of 0.15 dissociation would compare with the approximate solution. From the table the corresponding value is about 0.163, an error of about +8.5%.

Exact Enough Solution - Using Linear Series Solution

In case the hand-waving argument above didn't persuade you that the approximate solution is always bigger that the exact solution, let's explore a solution using a series where an increasing number of terms could be used to get a solution as exact as desired.

Rearranging equation (8), an exact equation, by pulling out a $\sqrt{\dfrac{K_d}{c_0}}$ term:

\begin{align*} \alpha &= \sqrt{\dfrac{K_d}{c_0}} \left( \sqrt{\dfrac{K_d}{4\cdot c_0} + 1} - \sqrt{\dfrac{K_d}{4\cdot c_0}} \right)\tag{9} \\ &= \sqrt{\dfrac{K_d}{c_0}} \left( \sqrt{\dfrac{K_d}{4\cdot c_0} + 1} - \sqrt{\dfrac{K_d}{4\cdot c_0}} \right)\cdot \left(\dfrac{\sqrt{\dfrac{K_d}{4\cdot c_0} + 1} + \sqrt{\dfrac{K_d}{4\cdot c_0}}}{\sqrt{\dfrac{K_d}{4\cdot c_0} + 1} + \sqrt{\dfrac{K_d}{4\cdot c_0}}} \right)\tag{10} \\ &=\dfrac{\sqrt{\dfrac{K_d}{c_0}}}{\sqrt{\dfrac{K_d}{4\cdot c_0} + 1} + \sqrt{\dfrac{K_d}{4\cdot c_0}}} \tag{11} \end{align*}

Now, using the Taylor series for the denominator yields an approximation, which is quite good. (Many thanks to user Claude Leibovici on the Mathematics site for the heavy lifting here...)

$${\sqrt{\dfrac{K_c/c_0}{4} + 1} +\sqrt{\dfrac{K_c/c_0}{4}} }=1+\frac{\sqrt{K_c/c_0}}{2}+\frac{K_c/c_0}{8}+O\left((K_c/c_0)^{2}\right)\tag{12}$$ which makes

$$\alpha^{"} \approx \frac {\sqrt{K_c/c_0}}{1+\frac{\sqrt{K_c/c_0}}{2}+\frac{K_c/c_0}{8}}\tag{13}$$

Obviously the more terms the better the approximation, but three terms is good enough for anything that we chemists would do. Again remember that we're solving for dilute solutions where $K_c/c_0 \ll 1$. Also notice that all the terms are positive so the more terms used, the smaller the value of the dissociation fraction.

\begin{array}{|c|c|c|c|c|} \hline K_c/c_0 & Quadratic & \sqrt{K_c/c_0} & \frac {\sqrt{K_c/c_0}}{1+\frac{\sqrt{K_c/c_0}}{2}} & \frac {\sqrt{K_c/c_0}}{1+\frac{\sqrt{K_c/c_0}}{2} + \frac{K_c/c_0}{8}} \\ \hline 1 & 0.61803399 & 1.00000000 & 0.66666667 & 0.61538462 \\ \hline 0.81 & 0.58192705 & 0.90000000 & 0.62068966 & 0.58017728 \\ \hline 0.64 & 0.54162637 & 0.80000000 & 0.57142857 & 0.54054054 \\ \hline 0.49 & 0.49663670 & 0.70000000 & 0.51851852 & 0.49601417 \\ \hline 0.36 & 0.44641839 & 0.60000000 & 0.46153846 & 0.44609665 \\ \hline 0.25 & 0.39038820 & 0.50000000 & 0.40000000 & 0.39024390 \\ \hline 0.16 & 0.32792156 & 0.40000000 & 0.33333333 & 0.32786885 \\ \hline 0.09 & 0.25835623 & 0.30000000 & 0.26086957 & 0.25834230 \\ \hline 0.04 & 0.18099751 & 0.20000000 & 0.18181818 & 0.18099548 \\ \hline 0.01 & 0.09512492 & 0.10000000 & 0.09523810 & 0.09512485 \\ \hline 0.0001 & 0.00995012 & 0.01000000 & 0.00995025 & 0.00995012 \\ \hline 0.000001 & 0.00099950 & 0.00100000 & 0.00099950 & 0.00099950 \\ \hline 0.00000001 & 0.00010000 & 0.00010000 & 0.00010000 & 0.00010000 \\ \hline 1E-10 & 0.00001000 & 0.00001000 & 0.00001000 & 0.00001000 \\ \hline 1E-12 & 0.00000100 & 0.00000100 & 0.00000100 & 0.00000100 \\ \hline \end{array}