3
$\begingroup$

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?

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4
  • $\begingroup$ I just edited the question to pinpoint what I think Limbo Productions was trying to ask. // User Limbo Productions if I missed interpreted your question please either edit the current version, or revert back to your original question. $\endgroup$
    – MaxW
    May 6, 2020 at 18:29
  • $\begingroup$ @MaxW The flaw with the question is that it doesn't state what is known beforehand, or what would be regarded as a tolerable error. You provide a nice explanation of how the approximate $\alpha$ provides an upper bound on the true $\alpha$, and that is certainly useful. But otherwise this question seems pointless. If you know the values of $K_d$ and $c_0$ and of the tolerated error, then it is trivially found if the approximation is good enough. Maybe MFarooq is onto something regarding the point of the question, but we'll probably never know (and honestly I don't understand his explanation). $\endgroup$
    – Buck Thorn
    May 7, 2020 at 16:05
  • $\begingroup$ @BuckThorn - I took the OP's question to mean "in general". So for whatever problem comes along The tolerable error and $K_d/c_0$ would be for a particular known. Then for that particular problem, how is the OP to decide if the approximation is good enough? // My take away from MFarooq's point was the total effort to (1) make the approximate calculation and (2) to do a check is almost as great as the effort to solve the quadratic which gives the exact solution. So why bother with the approximation? Just solve the quadratic and be done with it. $\endgroup$
    – MaxW
    May 7, 2020 at 17:40
  • $\begingroup$ @MaxW If that was MFarooq's point I completely agree. $\endgroup$
    – Buck Thorn
    May 7, 2020 at 17:46

4 Answers 4

4
$\begingroup$

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}

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4
  • $\begingroup$ I was hoping for a more rigid explanation rooted in mathematics but I guess it has to come down to comparing values and noticing when they start to get reasonably close to each other. Less satisfying but I will take it. Thank you for your help I appreciate it. $\endgroup$
    – Amadeus
    May 7, 2020 at 15:55
  • $\begingroup$ @l1mbo - I'm befuddled as to what else you might want. It is easy to calculate the exact error. Let $\alpha$ be the solution from the quadratic solution, and $\alpha'$ be the solution from the approximation. Then: $$\pu{\% Error} = 100\times (\alpha' -\alpha)/\alpha)$$ But obviously if you're going to calculate $\alpha$, which is exact, why bother calculating $\alpha'$? $\endgroup$
    – MaxW
    May 7, 2020 at 17:48
  • $\begingroup$ [I know this is 1.5 yrs late, I just happened to be going through my old posts on stack exchange so I stumbled on this, so I'll leave this here for anyone referring to this in the future] what I wanted was, in hindsight, extremely simple. The exact expression for $\alpha = - \dfrac{K_d}{2\cdot c_0} + \sqrt{\left(\dfrac{K_d}{2\cdot c_0}\right)^2 + \dfrac{K_d}{c_0}}\tag{4}$ can be rewritten as function only of $\alpha'$ by replacing $\dfrac{K_d}{c_0}=\alpha'^2$. $\endgroup$
    – Amadeus
    Sep 19, 2021 at 17:35
  • $\begingroup$ Then, we can replace all $\alpha$ in the % error $= 100\times (\alpha' -\alpha)/\alpha)$ and thus % error can be written entirely as a function of $\alpha'$. Finally when we graph the error function we can see at what $\alpha'$ the error becomes greater than whatever value we see fit (5% error here) imgur.com/a/LsX7YlS This was 'more rigid' explanation I sought. (No idea why I struggled to figure out this rather elementary answer then) $\endgroup$
    – Amadeus
    Sep 19, 2021 at 17:35
3
$\begingroup$

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.

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7
  • $\begingroup$ Hmmm... You could a search and iteratively solve for $\alpha$ with that equation. However in order to solve for $\alpha$ directly you need some function of $K_d/c_0$. It would seem that using the first three terms would give you some idea of the error. $\endgroup$
    – MaxW
    May 5, 2020 at 9:01
  • $\begingroup$ dah... Don't need the series. Let $\alpha = \sqrt{K_d/c_0}$ and compare $\alpha$ and $\alpha /\sqrt{1-\alpha}$ to get some idea of what the error is. It won't be exact, but in the ballpark. $\endgroup$
    – MaxW
    May 5, 2020 at 9:36
  • $\begingroup$ yep: that was what I thought. $\endgroup$
    – porphyrin
    May 5, 2020 at 10:43
  • $\begingroup$ @porphyrin, The student is talking about the "our formula derived with approximations will not give us the true value of the degree of dissociation" ($sic$), he is wondering about the validity of the formula itself, not its quadratic approximations. It is the definition of alpha that is the Achilles heel of this whole experiment in terms of conductance measurements. I know how this experiment is taught in South Asia so I can feel what he is trying to ask although quite vaguely. We used make the students solve the whole quadratic equation, which is not a big deal. $\endgroup$
    – AChem
    May 5, 2020 at 13:37
  • $\begingroup$ The question was about how you could tell if the dilute approximation was any good. If $\alpha = \sqrt{K_d/c_0}$ then comparing $\alpha$ and $\alpha/\sqrt{1-\alpha}$ gives a good indication. // Solving the quadratic will always give the exact answer of course. $\endgroup$
    – MaxW
    May 5, 2020 at 15:37
2
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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.

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4
  • $\begingroup$ Granted other ways may exist, but that doesn't mean we do not determine dissociation solely on the basis discussed above. $\endgroup$
    – Amadeus
    May 5, 2020 at 4:32
  • $\begingroup$ Equation (B) is the typical shortcut to an answer assuming that there is is little disassociation. If $\alpha^{'} \le 0.1$ the answer will be within 5%, if If $\alpha^{'} \le 0.010$ the answer will be within 1%. So you don't always have to solve the quadratic, though the quadratic will always give the exact answer. $\endgroup$
    – MaxW
    May 5, 2020 at 15:16
  • $\begingroup$ " only weak acids or bases can be measured by conductivity measurements." Isn't the point that you don't have to worry (as much) about activity coefficients in weakly dissociating electrolytes? In any case, I find the last paragraph confusing. $\endgroup$
    – Buck Thorn
    May 5, 2020 at 19:19
  • $\begingroup$ I was referring to the the determination of dissociation constants by conductivity measurements. Only relatively weak acids or bases can be employed for such experiments. I have tried to clarify the the last paragraph. $\endgroup$
    – AChem
    May 5, 2020 at 19:32
0
$\begingroup$

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}

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12
  • $\begingroup$ I am confused, what is the conceptual difference between equation (2) (3) and (8)? It seems like a circular argument. $\endgroup$
    – AChem
    May 5, 2020 at 1:13
  • $\begingroup$ @M.Farooq - It is the difference between equations 5 and 6 that is important. Note alpha and alpha-prime are calculated and that alpha-prime > alpha always. $\endgroup$
    – MaxW
    May 5, 2020 at 1:17
  • $\begingroup$ I still don't understand how the central question of why we think it's okay to use alpha- prime which we derive through an approximation instead of alpha to judge whether the approximation was correct in the first place $\endgroup$
    – Amadeus
    May 5, 2020 at 2:49
  • 1
    $\begingroup$ I think you changed the equation labels. My question remains: the final series expansion does not seem an advantage over the exact solution (it seems even more complicated to compute). And the advantage of the simplified formula is that you can quickly evaluate it in your calculator or even your head (back when people knew how to estimate square roots in their heads). The advantage is more or less null in our day of broadly available computing power. $\endgroup$
    – Buck Thorn
    May 6, 2020 at 13:04
  • 1
    $\begingroup$ @BuckThorn - I did number the last two equations, otherwise the equations have kept their numbering. In one of the revisions I did move the equations around, but I kept the numbering. // I added a second answer which is much more the the OP's question. My final point there was if you're going to use the approximation, then check the error bound, why not just use the quadratic to begin with since it gives the exact answer for a pittance of extra work. $\endgroup$
    – MaxW
    May 6, 2020 at 16:00

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