Why does the pH of a diluted solution change?

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SUMMARY

The pH of a diluted solution changes due to variations in hydronium ion concentration [H3O+], despite the Henderson–Hasselbalch equation suggesting otherwise under certain conditions. When applying the equation, the assumption that concentrations remain constant leads to inaccuracies. The common ion effect and Le Chatelier's principle indicate that weak electrolytes ionize less in the presence of strong electrolytes, affecting the equilibrium concentrations of acid and base. Accurate pH calculations should consider activities rather than formal concentrations, and tools like Borek's pH calculator utilize the Debye-Hückel theory for improved pH estimations in buffers and titrations.

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  • Understanding of the Henderson–Hasselbalch equation
  • Knowledge of weak and strong electrolytes
  • Familiarity with Le Chatelier's principle
  • Basic concepts of chemical activity and concentration
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  • Learn about the common ion effect in buffer solutions
  • Explore the Debye-Hückel theory for ionic solutions
  • Investigate the use of Borek's pH calculator for buffer and titration calculations
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Chemistry students, laboratory technicians, and professionals involved in analytical chemistry or buffer solution preparation will benefit from this discussion.

Bohrok
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The first time I came across a chemistry problem about the pH of a solution that's been diluted, I thought there was an argument for a change and for no change. After doing some searching online, it seems that the pH does change since [H3O+] changes, which I understand. However, using the Henderson–Hasselbalch equation equation both before and after dilution would give the same pH, right?

As an example, if I have x moles each of HA and A- in 1 L solution, thenpH = pK_a + \log\left(\frac{\frac{x\text{ mol }A^-}{1 L}}{\frac{x\text{ mol }HA}{1 L}}\right) = pK_a + \log(1) = pK_aNow if the solution were at 2 L to start with, thenpH = pK_a + \log\left(\frac{\frac{x\text{ mol }A^-}{2 L}}{\frac{x\text{ mol }HA}{2 L}}\right) = pK_a + \log(1) = pK_a

How is this setup giving the wrong answer as opposed to recalculating [H3O+] after the dilution?
 
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It looks like x moles/x moles=1 is the error. It should look more like X mol per L/Y mol per L or (.10 M/.12 M), when X=.10, Y=.12 and M=(mol/L).

Due to the common ion effect and Le Chatelier's principal, the weak electrolyte ionizes less than the strong electrolyte. Therefore the equilibrium shifts in favor of the strong electrolyte, creating an unequal molar concentration of acid and base (example .10 M and .12 M). Hopefully I got the question right.
 
A hint; take a look at the derivation of the HH equation and the simplifying assumptions that are made...

On the surface definition of a buffer, the pH is largely defined by pKa and has a broad largely unchanged pH over a range defined by the buffer's capacity. Dilution of the buffer only changes the capacity but not the pH range to the first order approximation.

In reality, the quantities used are not the formal concentrations of the corresponding acid and conjugate, but their Activities which do have a bearing on changing the pH as one changes concentration.

Borek's pH calculator (chembuddy) uses Debye Huckel and Extended Debye Huckel Limiting Law to make better approaches to an iterative solving of the pH in buffers and titrations. He has a good explanation on the web site.
 

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