We have performed the following redox titration

(1): $\ce{MnO_4^- + 8H^+ + 5Fe^2+ <=> Mn^2+ + 4H_2O + 5Fe^3+}$

where we have used a iron(II)solution (the analyte) and titrated it with a permanganate solution (titrant). The iron(II) was added by measuring up $\ce{FeSO_4*7H_2O}$. There is also $\ce{H_2SO_4}$ and $\ce{NaNO_3}$ in the analyte solution. The permanganate solution was prepared by an instructor. When the equivalence point was reached we could calculate the original concentration of $\ce{Fe^2+}$ in the analyte solution, we have pe-values, E-values, the volume of titrant needed to reach the equivalence point and a lot more different values.

The equilibrium constant lgK for

(2): $\ce{Fe^3+ + e- <=> Fe^2+}$

is 13,02 (from a data collection). Our calculated value for lg'K based of the concentrations that was calculated in the titration in reaction (2) is 11,67.

We know that $\ce{Fe^3+}$ forms a complex with $\ce{SO4^2-}$

(3): $\ce{Fe^3+ + nSO_4^2- <=> Fe(SO_4)_n^{3-2n}}$

where n=1 or 2. This is supposed to be a reason for the difference in the theoretical value of the equilibrium constant and our calculated one(the difference between 13,02 and 11,67). Why does this complex formation make our calculated value lower than the theoretical one? How does this affect the equilibrium constant?

Another reason for this difference is the ionic strength in the solution, but in which way? What is influencing the ionic strength in this system?

(there have been no previous calculations with ionic strength or activities in this laboration and there is no other mentioning of ionic strength anywhere else in our instructions)

The calculation of our lg'K has been calculated as following:

$\ce{'K = \frac{[Fe^2+]}{[Fe^3+]*({e^-)}}}$


$\ce{lg'K = lg[Fe^2+] - lg[Fe^3+] + pe}$

Where [] are the concentrations in Molar and () is the activity (we can not write $\ce{e^-}$ as a concentration but we have pe-values).

  • $\begingroup$ How are you calculating all these values and ultrafast reactions by titration (which technique)? $\endgroup$
    – ACR
    Commented Nov 23, 2022 at 19:20
  • $\begingroup$ There are missing relevant details, context and background of the question. // What time passed between formulation of your questions and posting them ? What topic research have you done before asking? $\endgroup$
    – Poutnik
    Commented Nov 23, 2022 at 19:57
  • $\begingroup$ I appreciation that you are thinking deeply. Okay, so you are using the Nernst equation. You have to check how the other author (person) got the value of 13.02? What is the citation and their experimenal condition? If those people employed different conditions, you cannot compare your values at all. $\endgroup$
    – ACR
    Commented Nov 23, 2022 at 22:28
  • $\begingroup$ Also, curious about the meaning lg'K? What is the purpose of the prime over the log function? $\endgroup$
    – ACR
    Commented Nov 23, 2022 at 22:29
  • $\begingroup$ The lgK value of 13,02 is a thermodynamic equilibrium constant, calculated at 25.0°C at 1 bar. It is calculated with the activities of the ions in solution in mind. Our lg'K of 11,67 is not calculated with the activities in mind. The temperature is therefore one reason for the deviation.The prime is only there to indicate that the activities is not included in the calculation, I guess you could also write "lgKc" to show that it is only calculated with the concentrations in mind. $\endgroup$
    – Tove
    Commented Nov 23, 2022 at 22:45

1 Answer 1


I suspect the primary source for the error (≈10%) you are encountering between your calculated $K'$ value and the theoretical $K$ value is the deviation of the molar equilibrium concentrations of $\ce{Fe^2+}$ and $\ce{Fe^3+}$ from the activities of both ions.

The presence of many ions in a solution, especially when some possess multiple charges (like $\ce{Fe^3+}$ and $\ce{Fe^2+}$, not to mention the sulphate iron complexes noted) or are highly concentrated, contribute to a non-negligible ionic strength $I_c$. The reason for this is the following:

$I_c$ for a solution with $n$ charged species can be calculated with the formula:

$$I_c=0.5\sum_{i=1}^n c_i\;z_i^2$$


$z_i$ represents the charge of species $i$

$c_i$ represents the concentration of species $i$

When $I_c$ is relatively high ($I_c>0.01M$), the activity coefficient $\gamma$ of multi-charged ions deviate significantly from unity (are reduced significantly), and the ones with a higher charge tend to experience a higher reduction than those with a lower charge.

The activity of a species can be calculated using its activity coefficient and concentration in solution:


This means that the relationship between $K$ and $K'$ in our case is:


Or simply:


We can observe that as both activity coefficients approach unity, $K$ approaches the value of $K'$:

$$\gamma_{\ce{Fe^3+}}≈\gamma_{\ce{Fe^2+}}≈1\implies K_\gamma ≈1\implies K≈K'$$

However, if $I_c$ is high enough to cause significant deviation, such deviation will be more impactful over $\gamma_{\ce{Fe^3+}}$ than it would over $\gamma_{\ce{Fe^2+}}$, since the former has a higher charge than the latter.

Just to give you an idea of the magnitude of error that arises when assuming ionic strength is negligible in cases when it shouldn't be, we will consider the theoretical activity coefficients for both ions at 25°C and a solution with an ionic strength of 0.01M:




This means that the thermodynamic equilibrium constant $K$ would be more than 150% higher than the calculated equilibrium constant $K'$ with molar concentrations.

If you want to estimate the value of both activity coefficients in accordance with your experiment, you can use the Debye-Hückel equation.

After independently substituting all the parameters for $\ce{Fe^3+}$ and $\ce{Fe^2+}$, the resulting expressions are:



Or you can combine them and get a function of $K_\gamma$ in terms of $I_c$:


Plotting $K_\gamma$ vs $I_c$ would give:

enter image description here

I suspect the ionic strength of your particular solution at equilibrium has a value of approximately $4\times10^{-4}M$ since:

$$I_c=4\times10^{-4}M\implies K_\gamma=1.12=\frac{K}{K'}=\frac{13.02}{11.67}$$

However, if there were a significant difference between the experimental temperature and the theoretical temperature, that would also factor in the error, since both equilibrium constants $K$ and $K'$ are temperature-dependent. Measurement errors and impurities can also contribute to the error.

  • $\begingroup$ Funny is, I have never seen molality based I either, until in the link. $\endgroup$
    – Poutnik
    Commented Nov 24, 2022 at 6:33
  • $\begingroup$ Me neither, apparently $I_m$ is more precise than $I_c$. $\endgroup$
    – Sam202
    Commented Nov 24, 2022 at 6:37
  • 1
    $\begingroup$ Perhaps, include the improved D-H equation (with W. link) in its symbolic form too, so the literal values do no look like magical numbers. A great answer, BTW. $\endgroup$
    – Poutnik
    Commented Nov 24, 2022 at 6:42
  • $\begingroup$ This was a great answer, I will apply this to our data and see how this affects the equilibrium constant. And yes, I forgot to mention that the temperature in the laboratory was 23.0°C, which will have an impact. $\endgroup$
    – Tove
    Commented Nov 24, 2022 at 10:23

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