# Chemistry - Calculation of Degree of Dissociation from Ostwald's Dilution Law

## Solution 1:

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:

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

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 $$\alpha'$$:

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

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

## Solution 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.

## Solution 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.